linear transformation.pdf

Chapter 3

Linear Transformations In your previous mathematics courses you undoubtedly studied real-valued functions of one or more variables. For example, when you discussed parabolas the function f ( f (x) =  x 2 appeared, or when you talked abut straight lines the function f ( f (x) = 2x  arose. In this chapter chapter we study functions of several variables, variables, that is, functions of vectors. Moreover, their values will be vectors rather than scalars. The particular transformations that we study also satisfy a “linearity” condition that will be made precise later.

3.1 3.1

Defini Definiti tion on and and Exam Exampl ples es

Before defining defining a linear transformation transformation we look at two examples. examples. The first is not a linear transformation and the second one is. 2

and let W  = R. Define Define f : f : V  →  W  by f ( f (x1 , x2 ) = x1 x2 . Thus Thus,, f   f   is a function defined on a vector space of dimension 2, with values values in a one-dimen one-dimensional sional space. space. The notation is highly suggestive; suggestive; that is, f : f : V  → W   W   indicates that f   f   does something to a vector in V   V   to get a vector in W . W . For exampl example, e, f (1 f (1,, −1) = −1, f (1 f (1,, 2) = 2, etc. etc. We will see later later that that this function does not satisfy the “linearity” condition and hence is not a linear transformation.  Example 1. Let V  Let  V  =

R

and W  = R3 . Defin Definee L : V  → W  by L(x1 , x2 ) = (x1 , x2  −  x1 , x2 ). Here Here the functi function on L   takes a vector in R2 and transforms it into a vector in R3 . For example, L example,  L(1 (1,, −1) = (1, (1, −2, −1) and L and  L(2 (2,, 6) = (2, (2 , 4, 6). This particular particular function function satisfies the linearity condition condition below, and so would would be called a linear transformatio transformation n from R2 to R3 .  Example Example 2. Let V  =

2

R

Definition 3.1. Let V  Let V  and W  and W  be  be two vector spaces. Let  L be  L  be a function defined

on V   V   with values in W . W . L  will be called a linear transformation if it satisfies the following two conditions: x  and  yy in V  1. L(x + y ) =  L(  L (x ) + L(y ), for any two vectors  x in  V .. 109

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LINEAR LINEAR TRANSFORMA TRANSFORMATIONS  TIONS 

x ) = cL x  in  V . 2. L(cx =  cL((x ), for any scalar c scalar  c  and vector x vector  x  V  . Let’s go back to Example 2 and verify that we did indeed have a linear transformation. For any x any  xx  = (x1 , x2 ) and y and  yy  = (y1 , y2 ), we have L(x + y ) = L[(  L [(x x1  + y1 , x2  + y2 )] = [x [x1  + y1 , (x2  + y2 ) − (x1  + y1 ), x2  + y2 ] = (x ( x1 , x2  − x1 , x2 ) + (y (y1 , y2  − y1 , y2 ) = L =  L((x ) + L(y ). Thus, Thus, condition condition 1 holds. Moreover Moreover x) =  L[( L(cx  L[(cx cx1 , cx2 )] = (cx (cx1 , cx2  − cx1 , cx2 ) =  c(  c (x1 , x2  − x1 , x2 ) = cL =  cL((x ) Since 1 and 2 hold, L   is a linear transformation from R2 to R3 . The The read reader er should now check that the function in Example 1 does not satisfy either of these two conditions. Example 3.   Define L Define  L :

3

R

→ R2 by L by L((x1 , x2 , x3 ) = (x ( x3  − x1 , x1  + x2 ).

a. Compute Compute L  L((e 1 ), L(e 2 ), and L and  L((e 3 ). b. Show L Show  L  is a linear transformati transformation. on. c. Show L Show  L((x1 , x2 , x3 ) = x1 L(e 1 ) + x2 L(e 2 ) + x3 L(e 3 ). a. L[(1, [(1, 0, 0)] = (−1, 1), 1), L[(0, [(0, 1, 0)] = (0, (0, 1), 1), L[(0, [(0, 0, 1)] = (1, (1, 0). 0). b. L(x + y ) = L =  L[( [(x x1  + y  +  y1 , x2  + y2, x3  + y3 )] = ((x (( x3  + y3 ) − (x1  + y1 ), (x1  + y1 ) + (x (x2  + y2 )) = (x ( x3  − x1 , x1  + x2 ) + (y ( y3  − y1 , y1  + y2 ) = L(  L (x ) + L(y ) x) = L L(cx =  L[( [(cx cx1 , cx2 , cx3 )] = (cx (cx3  − cx1 , cx1  + cx2 ) = c(  c (x3  − x1 , x1  + x2 ) =  cL(  cL (x ) Thus L Thus  L  satisfies conditions 1 and 2 of Definition 3.1, and it is a linear transformation. c. L(x1 , x2 , x3 ) = L(  L (x1e 1  + x2e 2 + x3e 3 ) = L(  L (x1e 1 ) + L(x2e 2 ) + L(x3e 3 ) = x1 L(e 1 ) + x2 L(e 2 ) + x3 L(e 3 ) Notice that c that  c  implies that once  L(  L (e k ), k ),  k  = 1, 2, 3, are known, the fact that  L  is x  in R3 .  a linear transformation completely determines  L(  L (x ) for any vector x vector  x We collect a few facts about linear transformations in the next theorem. Theorem 3.1. Let  L   be a linear transformation from a vector space  V   into a  vector space  W . Then 

00)) = 0 1. L(0

111

3.1. DEFINI DEFINITIO TION N AND EXAMPLES  EXAMPLES 

2. L(−x ) =  − L(x ) n

3. L

akx k  =

k=1

Proof.

n

  

ak L(x k )

k=1

00). 1. Let x Let  xx  be any vector in  V .  V  . Then L Then  L((x ) =  L(  L (x + 00)) =  L(  L (x ) + L(0 ). Adding Adding 00)) = 00,, where the zero vector on the −L(x) to both sides, we have L(0 left-hand left-hand side is in V  in  V  while the zero vector on the right-hand right-hand side is in W  in  W .. 00)) = L 2. 0 =  L(0  L(0 =  L((x − x ) =  L(  L (x ) + L(−x ). Thus L Thus  L((−x ) =  − L(x ). 3. We show that this formula is true for n  = 3 and leave the details of an induction argument to the reader. L(a1x 1  + a2x 2  + a3x 3 ) =  L(  L (a1x 1  + a2x 2 ) + L(a3x 3 ) =  L(  L (a1x1 ) + L(a2x2 ) + L(a3 x3 ) =  a 1 L(x 1 ) + a2 L(x 2 ) + a3 L(x 3 )

→ R4 be a linear transform transformation. ation. Suppose we know know that L(1, (1, 0, 1 ) = (−1, 1, 0, 2), L(0, (0, 1, 1) = (0, 6, −2, 0), 0), and and L(−1, 1, 1) = (4, (4, −2, 1, 0). Determine L Determine  L(1 (1,, 2, −1).

Example 4. Let L :

3

R

 L  form Solution.  The trick is to realize that the three vectors for which we know  L form a basis F  of  R3 . Thus, Thus, all we need to do is find the coordinat coordinates es of (1, (1 , 2, −1) with respect to F  to  F ,, and then use 3 of Theorem 3.1. The change of basis matrix P  below P  below is such that [x [x ]T   P [x ]T  F  =  P [ S .

 

−1

   

1 0 −1 1 P  = 0 1 1 1 1

 

0 −1 1 1 2 −1 = 1 −1 −1

Using this matrix to find the coordinates of (1 , 2, −1) with respect to F  to  F ,, we have

 

       

0 −1 1 T  [1, [1, 2, −1]T  =  P [1  P  [1, , 2 , 1] = 1 2 1 − − F  S  −1 −1 1 Thus

−3 1 2  = 6 −1 −4

(1, (1, 2, −1) =  − 3(1, 3(1, 0, 1) + 6(0 6(0, 1, 1)+ (−4)(−1, 1, 1) and L(1, (1, 2, −1) =  − 3L(1, (1, 0, 1)+ 6L 6L(0, (0, 1, 1)+ (−4)L 4)L(−1, 1, 1) =  − 3(−1, 1, 0, 2) + 6(0 6(0, 6, −2, 0) − 4(4, 4(4, −2, 1, 0) = ( −13 13,, 41 41,, −16 16,, −6)



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A standard method of defining a linear transformation from Rn to Rm is by matrix matrix multiplica multiplication. tion. Thus, Thus, if  x = (x1 , . . . , xn ) is any vector in Rn and xT . Then A = [ajk ] is an m  ×  n  matrix, define L(x ) = Ax Then L(x ) is an m  ×  1 m matrix that we think of as a vector in R . The various various properties properties of matrix matrix multiplication that were proved in Theorem 1.3 are just the statements that  L is a linear transformat transformation ion from Rn to Rm .





1 −1 2 Example Example 5. Let A Let  A =  = . If  L is  L  is the linear transformation defined by 4 1 3 A, compute the following: a. L(x1 , x2 , x3 )

b. L(1, (1, 0, 0), 0), L(0, (0, 1, 0), 0), L(0, (0, 0, 1)

L(x1 , x2 , x3 ) = =

 

1 −1 2 4 1 3

    x1 x2 x3

x1  − x2  + 2x 2 x3 4x1  + x2  + 3x 3 x3

L(1, (1, 0, 0) = (1, (1, 4)T 

L(0, (0, 1, 0) = (−1, 1)T 

L(0, (0, 0, 1) = (2, (2, 3)T 

The reader should note that  L(  L (e 1 ) is the first column of  A of  A,,  L(  L (e 2) is the second column of  A,  A , and L and  L((e 3) is the third column.  x, then L(e k ) will be the In general, if  A  is an m an  m × n  matrix and L(x ) =  Ax kth column of the matrix  A.  A .

′

′

(a , b ) r

(a, b)

r

θ

ψ

ψ

Figure 3.1

Until now we’ve thought of a linear transformation as an expression combining n ing n variables  variables to produce a vector in  R m . If we limit ourselves ourselves to this algebraic viewpoint we miss a fuller appreciation of linear transformations. For example, consider the mapping that rotates the points in the plane through an angle θ about the origin. Thus, if the point (a, ( a, b) is rotated through an angle θ  to the position (a (a , b ), it turns out that the formulas relating ( a , b ) to (a, (a, b) imply that this is a linear transformation. In fact (cf. Figure 3.1), setting ′

′

′

r  = (a2 + b2 )1/2 = [( [ (a )2 + (b (b )2 ]1/2 , ′

′

′

113

3.1. DEFINITION AND EXAMPLES 

we have that ′

a = r cos(θ + ψ) = r(cos θ cos ψ − sin θ sin ψ) = a cos θ − b sin θ b = r sin(θ + ψ) =  r(sin θ cos ψ + sin ψ cos θ) = a sin θ + b cos θ ′

Thus,

 

a cos θ  = sin θ b ′

′

− sin θ cos θ

  a b

 

a −b , where a 2 + b2 = 1, we b a can think of  A as defining a linear transformation from R2 to R2 that rotates the plane about the origin through an angle  θ, where cos θ = a, sin θ = b. Note that A T  = A 1 corresponds to a rotation of  − θ. In the succeeding pages we sometimes describe a linear transformation in a geometrical manner as well as algebraically, and the reader should try to visualize what the particular transformation is doing. Now whenever we see a matrix  A  of the form

−

Example 6.  Describe in geometrical terms the linear transformation defined

by the following matrices:

   

0 1 . This is a clockwise rotation of the plane about the origin −1 0 through 90 degrees.

a. A =

b. A =

2 0

0 1 3

T 

A[x1 , x2 ] =



 1 2x1 , x2 3



T 

This linear transformation stretches the vectors in the subspace  S [ee1 ] by a factor of 2 and at the same time compresses the vectors in the subspace S [e 2 ] by a factor of  31 . See Figure 3.2.

 

−1 0 . For this A, the pair (a, b) gets sent to the pair ( −a, b). 0 1 Hence this linear transformation reflects R2 through the x2   axis. See Figure 3.3. 

c. A =

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LINEAR TRANSFORMATIONS 

(3.3) L[(3.3)] = (6, 1)

Figure 3.2

(a, b)

(−a, b)

Figure 3.3

Problem Set 3.1 1. Let L(x1 , x2 , x3 ) =  x1  − x2  + x3 . a. Show that L  is a linear transformation from

3

R

to

R.

x) =  Ax xT  for every x x  in b. Find a 1 × 3 matrix A  such that L(x

3

R

.

c. Compute L(eek ) for k  = 1, 2, 3. xT  = 0}. d. Find a basis for the subspace K  =  {x : Ax 2. Let L be a linear transformation from L(ee2 ) = (0, 2), L(ee3 ) = (8, 1).

3

R

to R2 such that L(ee 1 ) = (−1, 6),

a. L(1, 2, −6) =? b. L(x1 , x2 , x3 ) =? x) = Ax xT . c. Find a matrix A  such that L(x 3. Let L be a linear transformation from R3 to R5 . Suppose that L(1, 0, 1) = (0, 1, 0, 2, 0), L(0, −1, 2) = (−1, 6, 2, 0, 1), and L(1, 1, 2) = (2, −3, 1, 4, 0). Notice that the three vectors for which we know  L  form a basis of  R3 . a. Compute L(eek ) for k  = 1, 2, 3. b. L(x1 , x2 , x3 ) =? x) = Ax xT . c. Find a matrix A  such that L(x

115

3.1. DEFINITION AND EXAMPLES 

4. For each of the following functions f  determine an appropriate  V  and W . Then decide if  f  is a linear transformation from V  to W . a. f (x1 , x2 ) = (x1 , 0, 1) b. f (x1 , x2 ) = (x1  − x2 , x1 ) c. f (x) = (x, x) d. f (x1 , x2 , x3 ) = (x1 , x2 , x2 , x3 , x3 , x1 ) 5. Let L : V  → W   be a linear transformation. Let K   be any subspace of  x): x  is any vector in K }. Show that L(K ) is a V . Define L(K ) =  { L(x subspace of  W . 6. Let L : V  → W  be a linear transformation. Let H  be any subspace of  W . x) is in H }. Show that L 1 (H ) is a subspace of  Define L 1 (H ) =  {x : L(x V . −

−

7. Show that the function defined in Example 1 is not a linear transformation. 8. Let L1 and L 2  both be linear transformations from V   into W . Let B = f k ) = L 2 (f  f k ) for {f k , k = 1, . . . , n}  be any basis of  V . Suppose that L1 (f  x) = L2 (x x) for every vector x x  in  V . each k. Show that L 1 (x x) = Ax xT , show that L(eek ) is the 9. Let A = [ajk ] be an m × n matrix. If  L(x kth column of  A. 10. Let S  =  cI 2 , be an arbitrary 2 × 2 scalar matrix. Describe the geometrical effect that the linear transformation  SxxT  has on R2 .

 

d1 0  is an arbitrary 2  × 2 diagonal matrix. Describe 0 d2 xT  for various values what happens to x  under the linear transformation Dx of  d j .

11. Suppose D =

12. If  D   is any invertible 2  ×  2 diagonal matrix, describe geometrically the effects of the linear transformations defined by the two matrices  D 1 and D 1 D. −

−

13. Let P n  be the vector space of all polynomials of degree at most  n. Define  p L( p  p) =  t 2 p  p for each p  p in P n . Then L can be thought of as a mapping from P n  to some vector space W . List some of these vector spaces, and then show that L  is a linear transformation for each of your  W ’s. 14. Let V  = C [0, 1], the vector space of continuous functions defined on [0,1]. f ] = a. Define L[f  1 V  to R .

´  1 0

f  f (t)dt. Show that L is a linear transformation from

´  t

f  f (s)ds, for each t in [0,1]. Show that T  is a linear b. Define T [f  f ](t) = 0  f  transformation from V  to V .

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CHAPTER 3.

LINEAR TRANSFORMATIONS 

15. Show that the operation of differentiation can be viewed as a linear transformation from P n to P n 1 . −

16. Let V  = C [0, 1]. f ](x) = f  a. Let L : V  → V   be defined by L[f  f (x)sin x. Is L   a linear transformation? f  b. Let L : V  →  V   be defined by L[f  f ](x) = sin x + f (x). Is L a linear transformation? 17. Let A  =

  2 1

−3 . Let V  = M 22 . 5

x) = x xA  (matrix multiplication). Compute a. Define L : V  →  V  by L(x L(ee j ) for j = 1, 2, 3, 4, where the e j ’s denote the standard basis vectors of  M 22 . b. Show that L  is a linear transformation. x) =  Ax x. c. Repeat parts a and b for  L : V  → V   defined by L(x

3.2

Matrix Representations

In the preceding section, matrices were used to define linear transformations. In this section we show that every linear transformation between two finitedimensional vector spaces can be represented by a matrix. Suppose first that  L is a linear transformation from Rn to Rm . To find a matrix that can be used to represent L  we do the following: let  {e k }, k  = 1, 2, . . . , n be the standard basis of  Rn . Then L(eek ) = (a1k , a2k , . . . , amk ) (3.1) for some constants a1k , a2k . etc. The subscript convention is important to remember when forming the matrix A, that will represent L. Thus, A = [ajk ] is an m × n matrix, and the entries in the  kth column of  A  are the coordinates of  L(eek ) with respect to the standard basis in  R m . Example 1. Let L :

3

R

→ R4 be defined by

L(x1 , x2 , x3 ) = (−6x2  + 2x3 , x1  − x2  + x3 , − x1  + x2  − 6x3 , 3x1  − x2  + 4x3 ) Then L(ee 1 ) =  L(1, 0, 0) = (0, 1, −1, 3) = (a11 , a21 , a31 , a41 ) L(ee 2 ) =  L(0, 1, 0) = (−6, −1, 1, −1) = (a12 , a22 , a32 , a42 ) L(ee 3 ) =  L(0, 0, 1) = (2, 1, −6, 4) = (a13 , a23 , a33 , a34 )

117

3.2. MATRIX REPRESENTATIONS 

Thus,

 

 

0 −6 2 1 −1 1 A = 1 −6 −1 3 −1 4

x). Let xx  = The next task is to show how to use this matrix in computing L(x n m (x1 , . . . , xn ). Then, assuming L : R → R , n

n

        

x) =  L L(x

xk e k  =

k=1

k=1

n

=

xk L(eek )

m

m

j=1

k=1

n

ajk e j  =

xk

ajk xk e j

j =1

k=1

x) with respect to the standard basis in Rm can be Note: The coordinates of  L(x xT , where [x1 , . . . , xn ] = [x x]S . found by computing the matrix product Ax  Example 2.  In Example 1 we had

L(x1 , x2 , x3 ) = (−6x2  + 2x3 , x1  − x2  + x3 , −x1  + x2  − 6x3 , 3x1  − x2  + 4x3 ) with matrix representation:

        

 

0 −6 2 1 −1 1 A = 1 −6 −1 3 −1 4 x , we have Computing Ax

 

0 −6 2 1 −1 1 −1 1 −6 3 −1 4

x1 x2 x3

−6x2  + 2x3 x1  − x2  + x3  = −x1  + x2  − 6x3 3x1  − x2  + 4x3

x  gives us the coordinates of  L(x x) in Thus, Ax

4

R

.

 



The preceding computations were based upon the vector spaces being Rn and Rm and using the standard bases. None of this is necessary in order for us to interpert L  as matrix multiplication. Let L : V  → W  be a linear transformation from V   into W . Let F  = {f k : k = 1, . . . , n}, and B = {g j : j = 1, 2, . . . , m}   be bases of  V  and W , f k ) as a linear combination of  respectively. Proceeding as before, we write L(f  the basis vectors in G. f k ) =  a 1kg 1  + a2kg 2  + · · · + amkg m L(f  m

=

 j =1

ajk g j

 

(3.2)

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CHAPTER 3.

LINEAR TRANSFORMATIONS 

f k )]T  Let A = [ajk ] be the m × n matrix whose kth column is [L(f  G , the  coordinates  f k ] with respect to the basis  G. This matrix A depends upon three things: of  L[f  1. The linear transformation L 2. The basis F  in V  3. The basis G  in  W  If we change either of the bases picked, the matrix representation A   will also change. The next calculation illustrates how to use the representation  A to calculate x). Let xx  be any vector in  V . Let[x x]F  = [x1 , . . . , xn ]. We the coordinates of  L(x x] with respect to G are given by the want to show that the coordinates of  L[x x]T  x matrix product A[x . Computing L[x ] we have F  n

n

        

x) = L L(x

xk f k  =

k=1

k=1

n

=

m

ajk g j

xk

j =1

k=1

=

f k ) xk L(f 

m

n

j=1

k=1

ajk xk g j

x) with respect to the basis  G This equation says that the  j th coordinate of  L(x n x]T  is k=1 ajk xk , but this is just the  j th row in the m × 1 matrix A[x F .   In other  x]   directly, but rather the coordinates  words, when using  A  we do not compute  L[x x ]  with respect to the basis  G, that is , of  L[x



x)]T  x ]T  [L(x G  = A[x F 

 

(3.3)

and W  = R3 . Define L : V  → W  by L(x1 , x2 ) = (x1 −  x 2 , x1 , x2 ). Let F  = {(1, 1), (−1, 1)}, and let G = {(1, 0, 1), (0, 1, 1), (1, 1, 0)}. Example 3. Let V  =

2

R

a. Find the matrix representation of  L using the standard bases in both V  and W . b. Find the matrix representation of  L using the standard basis in V  and the basis G  in  W . c. Find the matrix representation of  L   using the basis F  in standard basis in R3 .

2

R

d. Find the matrix representation of  L  using the bases F  and G.

and the

119

3.2. MATRIX REPRESENTATIONS 

Solution.

a. L(ee 1 ) =  L(1, 0) = (1, 1, 0) = ee1  + e 2 L(ee 2 ) =  L(0, 1) = (−1, 0, 1) =  −e 1  + e 3

   

1 −1 A = 1 0 0 1

b. L(ee 1 ) = (1, 1, 0) = 0(1, 0, 1) + 0(0, 1, 1) + (1, 1, 0) = 0gg 1  + 0gg 2  + g 3 L(ee 2 ) = (−1, 0, 1) = 0(1, 0, 1) + (0, 1, 1) − (1, 1, 0) = 0gg 1  + g 2  − g 3

   

0 0 1 A = 0 1 −1

f 1 ) =  L(1, 1) = (0, 1, 1) = ee2  + e 3 c. L(f  f 2 ) =  L(−1, 1) = (−2, −1, 1) =  − 2ee1  − e 2  + e 3 L(f 

   

0 −2 A = 1 −1 1 1 f 1 ) = 0gg 1  + g 2  + 0gg 3 d. L(f  f 2 ) = 0gg 1  + g 2  − 2gg 3 L(f 

   

0 0 1 A = 1 0 −2



It is clear from this example that the matrix representation of a linear transformation depends upon which bases are used in  V  and in W . If  V  and W  are the same vector spaces, then normally (in this text always) only one basis is used, rather than two different bases for the same vector space.

→ R2 be a linear transformation. Let F  = {(1, 6), (−2, 3)}. Suppose the matrix representation of  L  with respect to  F  is Example 4. Let L :

2

R



2 8 A = −1 −4 x) for any vector  xx  in Compute L(x

2

R



.

Solution. Let x = (x1 , x2 ) be any vector in

2

x) using the . To compute L(x x]F , the coordinates of  x  with respect to the basis F . matrix A we need to find [x The change of basis matrix  P  below gives the basis F  in terms of the standard basis 1 −2 P  = 6 3 R

 

120

CHAPTER 3.

Using P 

−1

LINEAR TRANSFORMATIONS 

x]T  we calculate [x F 

      

1 x1 3 2  = x2 15 −6 1 1 3x1  + 2x2 = − 6x1  + x2 15

x1 x2

−1

x]F  = P  [x

x) with respect to F  are Thus, the coordinates of  L(x





2 8 x]F  = A[x −1 −4





(3x1  + 2x2 )/15 (−6x1  + x2 )/15

1 −42x1 + 12x2 = 21x1  − 6x2 15



Thus x) = L(x = = =

   

   

1 1 f 1 + f 2 (−42x1 + 12x2 )f  (21x1  − 6x2 )f  15 15 1 1 (−42x1 + 12x2 )(1, 6) + (21x1  − 6x2 )(−2, 3) 15 15 6x2  − 21x1 {2(1, 6) − (−2, 3)} 15 6x2  − 21x1 (4, 9) 15

      



1 0 0 1 0 0 0 0 . Thus F  is the standard 0 0 0 0 1 0 0 1 −2 1 x) = Bxx. Find basis of  M 22 . Let B  = . Define L : M 22 →  M 22 by L(x 3 4 the matrix representation of  L  with respect to the standard basis  F  of  M 22 . Example 5. Let F  =

   

−2 1 f 1 ) =  Bf  f 1  = L(f  3 4 f 1  + 3f  f 3 =  − 2f  f 2 ) =  Bf  f 2  = L(f 

−2 1 3 4

f 2  + 3f  f 4 =  − 2f 

−2 1 f 3 ) =  Bf  f 3  = L(f  3 4 f 1  + 4f  f 3 = f  f 4 ) =  Bf  f 4  = L(f 

−2 1 3 4

f 2  + 4f  f 4 = f 

   

   

   

1 0 −2 0  = 0 0 3 0 0 1 0 −2  = 0 0 0 3 0 0 1 0  = 1 0 4 0 0 0 0 1  = 0 1 0 4

 

 

121

3.2. MATRIX REPRESENTATIONS 

Thus, the matrix representation of  L  is

 

0 −2 0 −2 3 0 0 3

1 0 4 0

0 1 0 4

 



In the following pages, when we say A, an m × n matrix, is a linear transformation or represents a linear transformation without specifically mentioning a basis or vector spaces, it is to be understood that  V  = Rn , W  = Rm , and that the standard bases in both  V  and W  are being used.

Problem Section 3.2 1. Let L(x1 , x2 ) = (3x1  + 6x2 , −2x1 + x2 ) a. Find the matrix representation of  L  using the standard bases. b. Find the matrix representation of  L   using the basis F  = {(−4, 1), (2, 3)}.

 

 

6 1 −4 0 2. Let L : R2 → R4 have matrix representation A = with respect 2 9 8 −3 to the standard bases. a. L(ee1 ) =?, L(ee2 ) =? b. L(−3, 7) =?

c. L(x1 , x2 ) =?

x) =  L(L(x x)), 3. Let L be a linear transformation from R2 into R2 . Define L2 (x 3 2 +1 n n x) =  L(L (x x)), and L x) =  L(L (x x)). Suppose L(x1 , x2 ) = (ax1 + L (x (x bx2 , cx1  + dx2 ). a. Find the matrix representation A of  L  with respect to the standard bases. b. Show that the matrix representation of  L 2 with respect to the standard bases is A2 , and in general the matrix representation of  L n with respect to the standard bases is  A n . c. What can you say if some basis other than the standard basis is used? 4. Let V  = R3 and let F  = {(1, 2, −3), (1, 0, 0), (0, 1, 0)}. Suppose that the matrix A  represents a linear transformation L : R3 → R3 with respect to the basis F , where 0 1 −2 0 A = 2 1 5 0 1 a. L(1, 2, −3) =?

 

b. L(1, 0, 0) =?

 

c. L(x1 , x2 , x3 ) =?

122

CHAPTER 3.

LINEAR TRANSFORMATIONS 

x) =  cx x, 5. Let V  be a vector space of dimension  n. Define L : V  → V  by L(x where c is any constant. Let F   be any basis of  V . What is the matrix representation of  L  with respect to this basis? 6. Let L 1 (x1 , x2 ) = (x1 − x2 , 2x1 +3x2 ) and let L 2 (x1 , x2 ) = (2x1 − 5x2 , 3x1 − x) =  L 1 (x x) + L2 (x x). x2 ). Define (L1  + L2 )(x a. Find the matrix representations A1  and A2  of L1  and L2 , respectively, with respect to the standard basis of  R 2 . b. Show that the matrix representation of  L 1  + L2 is A 1  + A2 . 7. Let L 1 and L 2  be two linear transformations mapping R2 into R2 . Let F  be any basis of  R2 . Let A1 and A2  be the matrix representations of  L1 and L2 , with respect to the basis F , respectively. Show that the matrix representation of  L 1  + L2  with respect to F  is A 1  + A2 . 8. Let L(x1 , x2 , x3 ) = (x2  + x3 , 6x1  − x2  + 3x3 , 2x1  + 3x2 − 7x3 , 2x1  + 6x3 ). a. Compute L[eek ] for k  = 1, 2, 3. b. Find the matrix representation A, of  L, with respect to the standard bases in R3 and R4 . x] for any vector x x  in R3 . c. Compute A[x 4

9. Define L :

R

→ R2 by L(x1 , x2 , x3 , x4 ) = (x2  + 2x3  + 3x4 , 2x1 − 6x4 ).

a. Compute L(eek ) for k  = 1, 2, 3, 4. b. Find the matrix representation A, of  L, with respect to the standard bases in R4 and R2 . x] for any vector x x  in R4 . c. Compute A[x 10. Let L be a linear transformation from R3 to R2 . Let F  =  { (1, 1, 1), (0, 1, 1), (1, 1, 0)}  =  {f 1 , f 2 , f 3 }. Let G  =  { (1, 2), (2, 3)} =  {g 1 , g 2 }. Suppose that 2 0 4 the matrix representation of  L with respect to these bases is . 1 −2 1





f k )]G  =? a. For k  = 1, 2, 3, [L(f  f k ) for k  = 1, 2, 3. b. Compute L(f  c. Find the matrix representation of  L using the standard basis S  in R3 and the basis G  in R2 . d. Find the matrix representation of  L   using the standard basis S  in both R3 and R2 . 11. Let L :

2

R

→ R2 . Let F  = {f 1 , f 2 }  be a basis of  R2 . Suppose that A =

  −2 0 0 3

is the matrix representation of  L   with respect to the basis F . What is f k ) for k  = 1, 2? L(f 

123

3.2. MATRIX REPRESENTATIONS 

12. Let L be the linear transformation that rotates the plane through an angle of  θ  degrees. Let A  be the matrix representation of  L. Then as we saw in Section 3.1 cos θ − sin θ A = sin θ cos θ





Find the matrix representations of  L2 , L3 , . . . , Ln. (Hint: What is L2 geometrically?) 13. Let L1 and L2  be two linear transformations from R2 to x ) = L 1 (L2 (x x)). composition of  L 1  with L 2 by L 1 ◦ L2 (x

2

R

. Define the

a. Show that the composition of two linear transformations is also a linear transformation. b. If  A1  and A2  are the matrix representations of  L1  and L2  with respect to the same basis, respectively, show that the matrix representation of the composition L 1  ◦ L2  is given by the matrix product  A 1 A2 . 14. Find the matrix representations for each of the following linear transformations with respect to the standard basis of the vector space in question: a. L : P n → P n  p polynomial p  p.

−1

by L( p  p) =  p p , i.e., L( p  p) is the derivative of the ′

 p . b. L : P n  →  P n by L( p  p) = p ′

c. L : P n  →  P n+2 by L( p  p) = t 2 p  p.

´  t

15. Define L[  p  p](t) p = 0  p  p(s)ds, for each t  in [0,1]. Then L : P n  →  P n+1 . Find a matrix representation for  L  using the standard bases. 16. If  A is an m × n matrix, we can think of  A as a linear transformation from n m R to R . What spaces are appropriate for each of the following matrices to be thought of as a linear transformation? a. AT  b. AT A c. AAT  17. Let L  be a linear transformation from a vector space  V  to a vector space x) = 0 for every vector x in V . What must any W . Suppose that L(x matrix representation of  L  equal? 2

  

18. Let V  = { j=1 (aj cos jx  + bj  sin jx) : aj and bj   arbitrary}. L : V  →  V  by 2

L

   2

(− ja j sin jx + jb j cos jx)

aj cos jx + bj  sin jx  =

j=1

Define

j=1

a. Find a basis F  for the vector space  V . b. Find the matrix representation A  of  L  with respect to your basis.

124

CHAPTER 3.

LINEAR TRANSFORMATIONS 

19. Using the same notation as in Example 5, define L : M 22 → M 22 by x) = x B. Find the matrix representation of  L   with respect to the L(x standard basis of  M 22 .

      

1 0 1 1 1 1 . Define L : M 22 → M 23 by 1 1 0 1 1 0 −1 3 4 x] = xB = x L[x . Using the standard basis in M 23   and the 2 0 1 basis G  in  M 22 , find the matrix representation of  L.

20. Let G =

0 1 1 1

x] =  Bxx. 21. Let B and G  be as in problem 20. Define  L : M 32  →  M 22 by L[x Using the standard basis in M 32  and the basis  G  in  M 22 , find the matrix representation of  L. 22. Let L : M 22  →  M 22  be defined by L

    a b c d

 =

a 0 0 0

a. Show that L  is a linear transformation. b. Find the matrix representation of  L  with respect to the standard basis of  M 22 . x] = Bx x(L[x x ] = x xB) c. Show that there is no 2 × 2 matrix B such that L[x for all xx  in  M 22 . f ](t) = 23. Let V  be the vector space in problem 18. For each f  in V   define L[f  ´  t  f (s)ds. Show that L   is a linear transformation from V  to V . Find 0 its matrix representation with respect to the basis found in problem 18. Show that the product of the matrix found in problem 18 with the matrix found in this problem equals  I 4 .

3.3

Kernel and Range of a Linear Transformation

For any linear transformation  L, mapping V   into W , there are two important subspaces associated with L. The first is a subspace of  V  called the kernel of  L; the second is a subspace of  W  called the range of  L. In this section we define these two subspaces and describe their relation to the solution set of a system of equations. Definition 3.2. Let L : V  → W . The kernel of  L   is the set of vectors x

x) = 0 . Letting ker(L) represent the kernel of  L, we have in V   for which L(x x ) = 0 }. ker(L) =  {x : L(x





2 −6 4 Example 1. Let A =  be the matrix representation of  L. Find the 1 −1 2 kernel K  of this linear transformation.

3.3. KERNEL AND RANGE OF A LINEAR TRANSFORMATION 

Solution.   Since A is a 2 × 3 matrix, A :

R

3

→

2

R

125

. We are asked to find those

x  = (x1 , x2 , x3 ) such that



2 x  = Ax 1

   

x1 2x1  − 6x2  + 4x3 0 x2  =  = x1  − x2  + 2x3 0 x3

−6 4 −1 2

 

x  is in the kernel of  A if and only if 2x1 − 6x2 + 4x3  = 0 =  x1 − x2 + 2x3 . Thus, x Hence K  =  {(x1 , x2 , x3 ): x1  + 2x3  = 0 =  x2 }.  This example demonstrates that the kernel is just the solution set of a homogeneous system of linear equations. We note that K  has dimension equal to 1 and that (−2, 0, 1) is a basis of  K . Definition 3.3. Let L   be a linear transformation mapping V   into W . The

w in W   such that L(x x) = w w , for some vector xx range of  L  is the set of vectors  w x) for some x x  in V  }. in V . Thus, Rg(L) = range(L) =  {w : w  = L(x Example 2.  Find the range of the linear transformation in Example 1.

w  in R2 such that → R2 , the range of  A consists of those w x  = w w  has a solution. The augmented matrix of the associated system is Ax Solution.  Since A :

3

R



2 −6 4 w1 1 −1 2 w2



It is clear that this system has a solution regardless of the values of  w 1 and w 2 , e.g., x 1  = (6w2  − w1 )/4, x 2  = (2w2  − w1 )/4, and x 3  = 0. Thus, Rg(L) = R2 .  Theorem 3.2. Let  L : V  → W   be a linear transformation. Then 

a. ker(L)  is a subspace of  V . b. Rg(L)  is a subspace of  W . 0 ) = 0 , we know that both the kernel and the range are nonempty. Proof.   Since L(0 Thus, to show that these two sets are subspaces we may use Theorem 2.6. Hence, x  + yy ) = L(x x) + L(yy ) = suppose that x and y   are in K   = ker(L). Then L(x 0 + 0 = 0 . Thus, K   is closed under vector addition. Now let a   be any x) = aL(x x) = a00 = 0 , and we see that K   is also closed scalar; then L(ax under scalar multiplication. This shows that K   is a subspace. To see that u  and vv  are any two vectors in Rg(L). Then Rg(L) is a subspace, suppose that u x) = uu and L(yy ) = vv . Then there are two vectors  xx and yy in V   such that L(x x + y ) =  L(x x) + L(yy ) = u u + v  and Rg(L) is closed under addition. Similarly L(x u = aL(x x) = L(ax x). Since Rg(L) is closed under if  a is any constant, then au vector addition and scalar multiplication, it is a subspace of  W . Consider a system of  m  linear equations in n  unknowns a11 x1  + · · · + a1n xn  = b 1 .......................... am1 x1  + · · · + amn xn  =  b m

(3.4)

126

CHAPTER 3.

LINEAR TRANSFORMATIONS 

x] =  Ax x, where Let L be the linear transformation from Rn to Rm defined by L[x A is the coefficient matrix [ajk ] of (3.4). Then the kernel of  L is just the solution set of the homogeneous system associated with (3.4). For xx  is in ker(L) if and x) = 00, but L(x x) = 00 if and only if  Ax x  = 0 . That is, xx  is in ker(L) if  only if  L(x and only if  x is a solution of (3.4) with bj  = 0 for j = 1, 2, . . . , m. The range of  L consists of those vectors  bb in Rm such that there is an  xx in Rn for which x) = bb, i.e., Ax x  = bb. That is, bb  is in the range of  L  if and only if (3.4) has a L(x solution. Example 3.   Consider the following system of equations:

3x4  = b 1 −x1  + 2x2  + 2x1  + 3x2  + 7x3  + 8x4  = b 2 4x1 − 2x2  + 6x3 = b 3

(3.5)

Find the kernel and range of the coefficient matrix of the above system of equations. Then determine the dimensions of these two subspaces. Solution.  The coefficient matrix  A  equals

 

 

2 0 3 −1 2 3 7 8 4 −2 6 0

and is row equivalent to the matrix

 

1 0 2 1 0 1 1 2 0 0 0 0

 

x  is a solution to the homogeneous system, i.e., xx  is in ker(A) if and only Thus, x if  x 2  =  − x3  − 2x4 and x 1  =  − 2x3 − x4 . Thus, ker(A) =  { (x1 , x2 , x3 , x4 ): x1  = −2x3 − x4 , x2  =  − x3 − 2x4 }. A basis for ker(A) is {(−2, −1, 1, 0), (−1, −2, 0, 1)}. Thus, dim(ker(A)) = 2. The augmented matrix of (3.5)

 

2 0 3 b1 −1 2 3 7 8 b2 4 −2 6 0 b3

is row equivalent to

 

 

 

b1 −1 2 0 3 0 1 1 2 (b2  + 2b1 )/7 0 0 0 0 (26b1 − 2b2 + 7b3 )/14

(3.5) has a solution if and only if  26b1 − 2b2  + 7b3  = 0

Thus, Rg(A) = {(b1 , b2 , b3 ) : 26b1 −  2b2 + 7b3 = 0}. A basis for Rg(A) is {(1, 13, 0), (7, 0, −26)}  and dim(Rg(A)) = 2. 

3.3. KERNEL AND RANGE OF A LINEAR TRANSFORMATION 

127

In the previous example A : R4 → R3 , dim(ker(A)) = 2, and dim(Rg(A)) = 2. It is not a coincidence that we have the following relationship: dim(ker(A))+ dim(Rg(A)) = dim(R4 ). Theorem 3.3. Let  L   be a linear transformation from  V  to W , where  V  is a   finite dimensional vector space. Then 

dim(ker(L)) + dim(Rg(L)) = dim(V )

(3.6)

Proof.   Let dim(V ) = n. Suppose that dim(ker(L)) = k, where we assume

first that 0   < k < n. Let x j , j = 1, . . . , k  be a basis for ker(L) and let y j ,  j = 1, . . . , n −  k   be a set of  n  −  k   linearly independent vectors in V   such x1 ), . . . , L(yy 1 ), that S  = {x 1 , . . . , xk , y 1 , . . . , y n k }  is a basis of  V . Then {L(x . . . , L(yy n k )}  is a spanning set of Rg(L). Since the xx j  are in ker(L), we have xj ) = 0 for j = 1, . . . , k. Thus, { L(yy 1 ), . . . , L(yy n k )}  must span Rg(L). We L(x wish to show that this set is linearly independent, and hence forms a basis for Rg(L). To this end suppose that −

−

−

c1 L(yy 1 ) + · · · + cn

y n−k ) = k L(y

−

0

Setting zz  =  c 1y 1  + · · · + cn ky n k , we have L(zz ) = 00. Thus, zz  is in ker(L) and there are constants  a j  such that −

−

a1x 1  + · · · + akx k = z  = c 1y 1  + · · · + cn

yn

−1

k

−

Since the set S  is linearly independent, every one of the constants must be zero. In particular c1 = c2 = · · · = cn k   = 0, and we conclude that the set {L(yy 1 ), . . . , L(yy n k )}  is a basis for Rg(L). Hence we have −

−

dim(ker(L)) + dim(Rg(L)) =  k + (n − k) = n = dim(V ) This equation remains to be verified in the two cases  k = 0 and k = n. We leave the details as an exercise for the reader In the next section we show how one can easily determine the dimension of  Rg(L). This technique coupled with the above theorem gives us an effective means of determining how large the solution space of a set of homogeneous linear equations is, and hence the size of the solution set for any system of  linear equations, homogeneous or not; cf. Theorem 1.10. Before starting the next section, we define several items. Definition 3.4. Let L : V  → W  be a linear transformation. L is said to be

x) =  L(yy ) implies that xx  = yy . one-to-one if  L(x Definition 3.5. Let L : V  → W  be a linear transformation. L is said to be

onto if Rg(L) =  W .

128

CHAPTER 3.

LINEAR TRANSFORMATIONS 

→ R2 have matrix representation A, where A  is given below. Show L  is onto but not one-to-one. Example 4. Let L :

3

R



1 2 3 A = 0 1 2



Solution.  To see that L  is not one-to-one we observe that the vector (1 , −2, 1)

is in ker(L); that is, L(1, −2, 1) = (0, 0) = L(0, 0, 0), but (1, −2, 1)   = (0, 0, 0). Hence, L is not one-to-one. To see that  L is onto we have to show that Rg( L) = 2 2 x  = (x1 , x2 , x3 ) R . Thus, let (w1 , w2 ) be any vector in R . Our task is to find x x) = (w1 , w2 ). Equivalently we need to solve the following system such that L(x of equations: x1  + 2x2  + 3x3  = w 1 x2  + 2x3  = w 2 A solution to this system is x 1  =  w1  − 2w2 , x 2  = w 2 , and x 3  = 0.



Theorem 3.4. Let  L : V  →  W   be a linear transformation. Assume, for parts  b and c, that  V  and  W   are finite dimensional. Then 

a. L  is one-to-one if and only if  ker(L) =  {0 }. b. L  is one-to-one if and only if  dim(V ) = dim(Rg(L)). c. L   is onto if and only if  dim(Rg(L)) = dim(W ). Proof.   Suppose L   is one-to-one. We want to show that K   = ker(L) = {0 }.

x) = 0 =  L(0 0 ) and we must have x x  = 00. Thus, suppose that xx  is in K . Then L(x x) = L(yy ), we must have L(x x − y ) = 0. Conversely, suppose K  =  {0 }. Then if  L(x Hence, x  −  yy   is in K , and we conclude that x = y . Part b of this theorem is an easy consequence of part a and Theorem 3.3. Suppose that L   is oneto-one. Then we have dim(V ) = dim(Rg(L))+ dim(ker(L)) = dim(Rg(L))+ 0 = dim(Rg(L)). Conversely, if dim(Rg(L)) = dim(V ), then dim(ker(L)) = 0, and we have ker(L) =  { 0}. The verification of part c is left to the reader as an exercise.

Problem Set 3.3 1. Each of the matrices below represents a linear transformation from Rn to m R . Determine the values of  n and m for each matrix. Then determine their kernels and ranges and find a basis for each of these subspaces.





a. 1 0 1

b.

  1 0 1

c.

  1 1 0 1

d.

 

1 1 1 1

 

1 1 0 1 1 1 0 −1

3.3. KERNEL AND RANGE OF A LINEAR TRANSFORMATION 

129

2. Each of the matrices below represents a linear transformation from Rn to m R . Determine the values of  n and m for each matrix, and their respective kernels and ranges.

 

a. 1 2

b.

    1 2 −1 0 0 1

c.

 0 1

3. Let A be the coefficient matrix of the system of equations below. If  L is the linear transformation defined by  A, what is the range of  L  and what is its kernel? Does this particular equation have a solution; i.e., is ( −2, 1) in the range of  L? 2x1  − 5x2  + 3x3  =  − 2 x1  + 3x2 = 1 4. For each of the matrices below determine the dimension of its range and the dimension of its kernel. Then decide if the linear transformations represented by these matrices are onto and/or one-to-one.

 

a. 1 2

b.

    1 2 −1 0 0 1

c.

 1 2

5. For each of the matrices below determine the dimensions of their range and kernel. Then determine if the linear transformations represented by these matrices are onto and/or one-to-one.

a.

 

1 0 1 0

 

0 −1 0 4 0 0 0 1

b.

  

1 2 1 −1 0 1 1 0 1 1

  

3 −1 1 −1 0 1 1 0 0 0

6. Verify part c of Theorem 3.4. Remember, the range of  L   is always a subspace of  W . 7. Let L : V  →  W  be a linear transformation. Let  {x j : j = 1, . . . , n}  be a xj ): j = 1, . . . , n}  is a spanning set for basis of  V . Show that the set  { L(x Rg(L). 8. Let L :

Rn

→ Rm be a linear transformation.

a. Show that if  n > m, then L   must have a nontrivial kernel, i.e., dim(ker(L)) > 0. b. If  n  ≤  m does L  have to be one-to-one? 9. Let L :

n R

→ Rm be a linear transformation.

a. If  n < m, show that L  cannot be onto. b. If  n  ≥  m, must L  be onto?

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10. Let L : V  →  W   be a linear transformation. Let Q be that subset of  W  that contains all vectors in  W  not in the range of  L, i.e., Q  = W \Rg(L). Is Q  a subspace of  W ? 11. Let S   be any n  ×  n   scalar matrix, i.e., S  = cI n   for some constant c. Determine the kernel and range of  S  for various values of  c. 12. Let D  be any n × n diagonal matrix. Determine the kernel and range of  D for various values of the diagonal entries. For example, what happens if the entry in the 1,1 position of  D  is zero? Is nonzero? 13. Characterize the kernel and range for the linear transformations in problems 13, 14, and 15 in Problem Set 3.1. 14. For each of the matrices A in problem 1, compute AT . Then determine the kernel and range of  A T . 15. For each of the matrices A   in problem 1, compute AT A. Determine if  these product matrices are one-to-one and/or onto. Compare the kernels of  A T A and  A. 16. Let L : Rn → Rm be a linear transformation. Show that if  L is both onto and one-to-one, then  m = n. 17. Verify Theorem 3.3 for the two cases ker(L) =  {0}  and ker(L) = V .

´  t

18. Let L : P 2 →  P 3  be defined by L[  p  p](t) p = 0  p  p(s)ds. Find a matrix representation for L using the standard basis in each of the vector spaces. Find a basis for the range and kernel of  L.  p (t). Find a matrix representa19. Let L : P 3  →  P 3  be defined by L[  p  p](t) p = p tion for L  using the standard basis in each of the vector spaces. Find a basis for the range and kernel of  L. ′





−1 2 5 20. Let B  = . 2 3 1 x] = x xB for any xx  in M 22 . Then L  is a linear transformation a. Let L[x from M 22 to M 23 . Find a basis for the kernel of  L and also a basis for the range of  L. x] = Bxx  for any xx  in M 32 . Then L  is a linear transformation b. Let L[x from M 32 to M 22 . Find a basis for the kernel of  L and also a basis for the range of  L. c. Find a matrix representation for each of the above two linear transformations. Use the standard basis.

131

3.4. RANK OF A MATRIX 

3.4

Rank of a Matrix

In the last section we proved a theorem that said the dimensions of the kernel, range, and domain of a linear transformation are related by the equation dim(V ) = dim(ker) + dim(Rg). We have also seen that the kernel is just the solution set for a system of homogeneous equations. In this section we show that the dimension of the range of  L is the same as the maximum number of  linearly independent rows or columns in a matrix representation of  L. Since this number is easy to calculate, we have an efficient method for computing the dimension of Rg(L) and hence an efficient method of determining the size of the solution set of a system of linear equations. Definition 3.6. Let A = [ajk ] be an m  ×  n   matrix. Each row of  A   can be

thought of as a vector in Rn and each column of  A  can be considered a vector in Rm . The row space of  A  is that subspace of  Rn spanned by the row vectors of  A, and the column space of  A is that subspace of  Rm spanned by the column vectors of  A.

Definition 3.7. Let A  = [ajk ] be an m × n matrix. The row rank of  A  is the

dimension of the row space of  A  and the column rank of  A  is the dimension of  the column space of  A.

 

 

−2 −1 1 1 . The row space of  A  is that subspace of  Example 1. Let A = 0 4 2 R spanned by the set {(−2, −1), (1,1), (0,4)}. Clearly this set spans a subspace of  R2 of dimension 2. Hence the row space of  A is R2 , and the row rank is 2. The column space of  A  is that subspace of  R3 spanned by the set {(−2, 1, 0), (−1, 1, 4)}. Since this set is linearly independent the column space has dimension 2. Thus the column rank is 2.  The fact that the row rank of  A  was equal to its column rank was no accident, as the following theorem shows. Theorem 3.5. Let  A = [aik ]  be an  m × n  matrix. Then the column rank and  the row rank of  A   are equal.

Proof.   Suppose the column rank of  A is p. Then 0 ≤ p ≤ n. If  p = 0, every

column of  A  is the zero vector in Rm , and hence every row of  A   is the zero vector in Rn . Thus the row space is the zero vector and we have row rank equal to 0 also. Now assume that p >  0. Let {z j : j = 1, . . . , p}  be a basis for the column space of  A, where z j = (z1j , z2j , . . . , zmj ) Then if  C k   is the kth column of  A, i.e., C k = (a1k , a2k , . . . , amk )T , there are constants b jk , 1 ≤  j  ≤  p, such that  p

C k  =

 j =1

bjk z T  j

132

CHAPTER 3.

LINEAR TRANSFORMATIONS 

Equating components, we have the following:  p

aik =



1  ≤  k  ≤  n, 1  ≤  i  ≤  m

bjk zij

j =1

Thus if  r i  is the ith row of  A, we have r i  = (ai1 , ai2 , ai3 , . . . , aij )

     p

=

 p

bj 1 zij ,

j=1

 p

bj 2 zij , . . . ,

j=1

bjn zij

j=1

 p

=

 

zij (bj1 , bj 2 , . . . , bjn )

j =1

Thus, the p row vectors (bj1 , . . . , bjn ), 1  ≤ j ≤ p, form a spanning set for the row space of  A. Hence, row rank ≤ p. This shows that the row rank of any matrix must be less than or equal to its column rank. Since the row rank of  A T  is the column rank of  A, and the column rank of  A T  is the row rank of  A, we also have that the column rank is less than or equal to the row rank. Hence the row and column ranks are equal. Definition 3.8.   The rank of a matrix is the common value of its row and

column rank. Example 2.  Compute the rank of the following matrix:

 

 

 

 

1 0 1 0 A = 2 1 1 1 0 2 3 2

Solution.  This matrix is easily seen to be row equivalent to the matrix.

1 0 1 0 B = 0 1 −1 1 0 0 5 0

This last matrix has rank equal to 3. Since the rows of  B   were obtained from linear combinations of the rows of  A and we can also obtain the rows of  A  from linear combinations of the rows of  B, the row spaces of  A and B must be the same. Hence, A and B have the same row rank and thus the same rank, namely,  3. In the preceding example we computed the row rank of  A by first finding the row rank of a matrix that was row equivalent to  A  and then used the fact that their row ranks were equal. We formalize this in the next theorem. Theorem 3.6. If  A  and  B  are two matrices that are row or column equivalent, then the rank of  A  is equal to the rank of  B .

133

3.4. RANK OF A MATRIX 

Proof.  See problem 10 at the end of this section.

We now need to relate these ideas to the problem of describing the solution x  = b , space of a system of equations. Consider the linear system of equations Ax where A = [ajk ] is an m×n matrix. The matrix A defines a linear transformation L from Rn to Rm . Since L(eek ) equals the k th column of  A, it is clear that the C k  be the kth column space of  A is the same as the range of  L. In fact if we let C  x  = (x1 , . . . , xn )T , then Ax x  = x1C 1 + · · · + xnC n ; that is, Ax x column of  A, and x is just a linear combination of the columns of  A. Thus, if  A  is



2 4



1 3 −2 8

we may write A[x1 , x2 , x3 ]T  as



2 4

        

1 3 −2 8

x1 2 x2  = x1 + x2 4 x3

1 3 + x3 8 −2

Clearly this is a linear combination of the columns of  A. These remarks make it clear that the column rank of  A (the dimension of  the column space) is equal to the dimension of the range of  L. Hence we have dim(ker(L)) =  n − dim(Rg(L)) =  n − rank of  A Remembering that ker(L) is just the solution space of the homogeneous system x  = 00, we have our promised algorithm for computing the size of the solution Ax space. What can be said about nonhomogeneous equations? Essentially the same thing, as the following theorem indicates. x0  = bb . Then the solution space of  Ax x  = bb Theorem 3.7.   Suppose  x 0  satisfies  Ax x0  + ker(A);   that is, every solution of  Ax x = bb   is equal to x 0   plus some  equals  x vector in the kernel. Note that this is just Theorem 1.10 restated. x  = bb. Then Proof.   Suppose Ax x − Ax x0  =  A(x x − x 0) 0 = bb − b  = Ax x0 + (x x − x 0 ). Conversely if  x  = x x0 +zz Thus x − x 0  is in the kernel of  A, and x  = x for some zz  in the kernel, then x  = A(x x0  + z ) =  Ax x0  + Azz  = bb + 0 = bb Ax Example 3.  Describe the solution set of the following system of equations:

2x1 + 4x3  + 5x4  = 8 x1  + 2x2 + 5x4  = 4 x1  + 6x2  + 3x3  + 10x4  = 11

134

CHAPTER 3.

LINEAR TRANSFORMATIONS 

Solution.  The coefficient matrix  A, which equals

is row equivalent to the matrix

 

2 0 4 5 1 2 0 5 1 6 3 10

 

1 0 0

5 2 5 4

0 1 0 0 0 1 0

 

 

Hence, dim(Rg(A)) = rank of  A  = 3. Since 4 − 3 = 1, dim(ker A) = 1. A basis x  = for the kernel is ( − 52 , − 45 , 0, 1). Thus, the solution space of the equation Ax x  = x x0  + c(− 25 , − 54 , 0, 1), for any constant c, assuming (8, 4, 11)T  is of the form x x0 . There is a solution if  of course that there is at least one particular solution x and only if (8,4,11) is in the range of  A and equivalently if and only if (8, 4, 11)T  is in the column space of  A. A basis of the column space is  { (2, 1, 1)T , (0, 2, 6)T , (4, 0, 3)T }. One easily sees that (8, 4, 11)T  = 2(2, 1, 1)T  + (0, 2, 6)T  + (4, 0, 3)T ; that is, bb  equals 2(first column) + (second column) + (third column) + 0(fourth column). Thus, xx0 = (2, 1, 1, 0)T  is a particular solution, and every solution is of the form 5 5 x  = (2, 1, 1, 0) + c − , − , 0, 1  2 4





A common problem has to do with data fitting. In its simplest form, this problem can be viewed in the following manner. Assume we have a collection of  n   points (xk , yk ), 1 ≤ k ≤ n, where xj  = xk if  j  = k. We wish to find a  p  p polynomial p  p(t) of degree m such that p  p(xk ) =  y k  for each k. In fact, we would like to find the smallest value of  m  such that regardless of the values  y k  such a polynomial will exist. Since there are n  data points and a polynomial of degree m has  m + 1 arbitrary coefficients we might conjecture  m  equal to n − 1 is the smallest value of  m for which we are guaranteed a solution. We now recast this problem in terms of linear transformations. Thus, let (xj , yj ), j = 1, 2, . . . , n,  p be given. Let p  p(t) be any polynomial in P m . Define L : P m  → Rn by L( p  p) = ( p  p(x1 ), p  p(x p 2 ), . . . , p  p(x p n )) That is, we evaluate our polynomial at the n  numbers xj . For example, if we  p had the three points ( −1, −2), (0,6), and (1,0) and p  p(t) = 8+ 2t − 8t3 + t4 , then  p L( p  p) = ( p  p(−1), p  p(0), p  p(1)) p = (15, 8, 3) The question then, as to whether or not a polynomial  p p   of degree m   can be  p picked so that p  p(xj ) = y j , amounts to deciding if the point (y1 , y2 , . . . , yn ) lies in the range of the linear transformation L. We next find a matrix representation

135

3.4. RANK OF A MATRIX 

for this linear transformation using the natural basis  {1, t , t2 , . . . , tm }  in P m  and the standard basis  {e 1 , . . . , en }  in Rn . L(1) = (1, 1, . . . , 1) L(t) = (x1 , x2 , . . . , xn ) L(tk ) = (xk1 , . . . , xkn )

k = 1, 2, . . . , m

Thus, we have the n × (m + 1) matrix 1 x1 x21   . . . xm 1 1 x2 x22   . . . xm 2 A = ..................... 1 xn x2n   . . . xm n

 

 

If all the xj ’s are different, it can be shown that the rank of  A  is the smaller of the two numbers m + 1 and n. Thus, if we wish to always be able to solve L( p  p) = (y1 , . . . , yn ) we need rank  A  =  n. Clearly, the smallest value of  m  that works is m = n −  1; i.e., A   is a square matrix. Another way of stating this is that for any  n  distinct numbers x 1 , . . . , xn  the linear transformation  L  maps P n 1  (polynomials of degree  n − 1 or less) onto Rn in a one-to-one fashion. −

Example 4.  Given the three points (1, −6), (2,0), and (3,6), we know from the

above discussion that there is a polynomial  p  p(t), of degree 3 − 1 = 2, such that  p  p  p  p(1) =  − 6, p  p(2) = 0, and p  p(3) = 6. Find this polynomial. Solution.   The transformation L : P 2 →

3

R

defined by the abscissas of these

three points satisfies 1 ) = (1, 1, 1) L(1

L(tt) = (1, 2, 3)

L(tt2 ) = (1, 4, 9)

The matrix representation for this transformation is

 

 

1 1 1 A = 1 2 4 1 3 9

 p This matrix has rank equal to 3. We wish to find p  p(t) such that L( p  p) = ( −6, 0, 6). 2  p In terms of the coefficients aj of   p(t) p = a 0  + a1 t + a2 t , we want a solution to the equation a0 −6 A a1  = 0 a2 6

     

The unique solution to this equation is a0 = −12, a1 = 6, a2   = 0. Thus,  p  p(t) =  − 12 + 6t is the unique polynomial of degree 2 or less that fits the data. We know that the polynomial is unique, since the matrix A has rank 3, which implies that dim(ker(L)) = 3 − 3 = 0. See Figure 3.4. 

136

CHAPTER 3.

LINEAR TRANSFORMATIONS 

(3,6)  p(t) = −12 + 6t

(2,0)

(1,−6)

Figure 3.4

Problem Set 3.4 1. Calculate the rank of each of the following matrices:





a. 1 0 1

b.

  1 0 1

c.

  1 1 0 1

d.

 

1 1 1 1

2. Calculate the rank of each of the following matrices:

 

a. 1 2

b.

    1 2 −1 0 0 1

c.

 

1 1 0 1 1 1 0 −1

 7 5

3. Each of the matrices below represents a linear transformation from Rn to m R . Determine the values of  n and m for each matrix. Then determine the dimensions of the range and kernel of  L

a.

 

1 0 1 0

 

0 −1 0 4 0 0 0 1

b.

  

1 1 0 1 1

2 −1 3 1 −1 −1 1 0 1 0 1 0 1 0 0

  

137

3.4. RANK OF A MATRIX 

4. For each matrix below, determine the dimensions of the range and kernel. Then decide if the linear transformation it represents is onto and/or oneto-one.

 

a. 1 2

b.

    1 2 −1 0 0 1

c.

 7 5

5. For each matrix below determine the dimensions of the range and kernel. Then decide if the linear transformation it represents is onto and/or oneto-one.

a.

 

1 5 2 1

 

−1 4 0 1

1 0 1 1

b.

  

  

2 3 1 −1 5 0 1 2 1 1

5 7 1 −1 5 1 3 4 0 0

6. Compute the row rank and column rank of each of the following matrices:





1 6 0 3 a. 2 −1 1 0

b.

 

−3 6 4 1 2 8 4 3 −4 1 0 0

 

c.

 

7. Consider the following system of linear equations: 2x1

1 2 4 5 8 −1 6 1

1 6 2 0

 

− 6x3  =  − 6 x2  + x3  = 1

Let A  be the coefficient matrix of this system. a. Compute the rank of  A. b. dim(ker(A)) =? Find a basis for ker(A). c. dim(Rg(A)) =? Find a basis for Rg(A). d. Is A  a one-to-one linear transformation? e. Is A  onto? f. Does the above system of equations have a solution? If yes, characterize the solution set in terms of the kernel of  A  and one particular solution. 8. Let L :

3

R

→ R2 have the matrix representation



1 3 −2 A = 4 −1 0



Show that the range of  L and the column space of  A are the same subspace of  R2 .

138

CHAPTER 3.

LINEAR TRANSFORMATIONS 

9. Consider the following system of linear equations: 4x1 + 2x3 + x4  = 0 2x1  − x2  + x3  + 3x4  = 1 −8x1  − 2x2  − 4x3 + 3x4  = 2 Let A  be the coefficient matrix of this system. a. Compute the rank of  A. b. dim(ker(A)) =? Find a basis for ker(A). c. dim(Rg(A)) =? Find a basis for Rg(A). d. Is A  a one-to-one linear transformation? e. Is A  onto? f. Does the above system of equations have a solution? If yes, characterize the solution set in terms of the kernel of  A and a particular solution. x1 , . . ., 10. Let xxk for k  = 1, 2, . . . , p be vectors in a vector space. Let  V  = S [x x p ]. If  c  is any nonzero constant show that x1 , x 2 , . . . , x p ]. a. V  = S [cx x1  + x 2 , . . . , x p ]. b. V  = S [x 1 , cx The reader should note that the result proved in this problem is what is needed to verify Theorem 3.6, the vectors then representing the rows or columns of a matrix. 11. Given data points (1,0) and (2,1), define L : P 2 → R2 by L( p  p) = ( p  p(1),  p  p  p  p(2)). Here p  p = p  p(t) is any polynomial of degree 2 or less. Find the matrix representation of  L with respect to the standard bases in  P 2  and R2 . Show that the rank of  A equals 2. What does this say about fitting polynomials of degree at most 2 through the data points (1,0) and (2,1)?  p be any polynomial in P 2 . Define L : P 2  → R3 by L( p 12. Let p  p) = ( p  p(−2), p  p(0), p  p  p(1)). Find the matrix representation  A of  L with respect to the standard bases in P 2 and R3 . Show that A has rank equal to 3. What does this say about fitting polynomials of degree 2 or less through three points in the plane with x  coordinates  − 2, 0, and 1? 13. Find a polynomial in  P 1 , if possible, that fits the following data: a (2,6), (3,6)

b. (2,0), (−1, 4)

c. (2,6), (3,6), (4,7)

14. Find a polynomial in  P 2 , if possible, that fits the following data: a. (−2, 6), (3,7) b. (−2, 6), (3,7), (4,7) c. (−2, 6), (3,7), (4,7), (5,8)

139

3.4. RANK OF A MATRIX 

15. Let V  = C [0, 1], the vector space of real-valued continuous functions defined on [0,1]. Define the mapping L  by





f (0) 1 f ] = L[f  + f  2 2

f   f  f (1) + 2

  1 2

f ] = [(sin0)/2 + sin 21  + (sin 1)/2]( 12 ). Thus, if  f  f (1) = sin t, we have L[f  This formula is just the trapezoid rule for the approximate evaluation of  integrals over the interval [0,1] using the points 0, 21 , and 1. a. Show that L is a linear transformation, after deciding of course what W  should be. Characterize the range and kernel of  L.

´  1

f ] = L[f  f ] − 0 f  b. Let L 1 [f  f (t)dt. Show that L 1  is also a linear transformation. How would you describe its kernel? 16. For each of the following matrices A   compute the rank of  A, AT , AAT , and A T A. You should get rank A T A equals rank A and rank AA T  equals rank A T . Hence all four numbers are equal.



a. 1 1 2





1 0 2 b. 0 1 3



c.

  1 3

−2 2

17. Let A  be any m × n matrix. How are the row space of  A  and the range of  AT  related? 18. Let x 1 , x2 , and x 3  be three different numbers. a. Show rank

      1 x1  = 2. 1 x2

1 x1 b. Show rank 1 x2 1 x3

x21 x22  = 3. x23





1 x1 Hint: The matrix in a is row equivalent to . 0 x2  − x1 19. Define L : P 2  → R3 by L[  p  p] p = ( p  p(x1 ), p  p(x p 2 ), p  p(x p 3 )), where the  x j ’s are all different. Show that L  is one-to-one and onto. 20. Let x 1 , x2 , . . . , xn be n  pairwise distinct numbers. Let 1 x1 x21   . . . xn1 1 1 x2 x22   . . . xn2 1 A = ...................... 1 xn x2n   . . . xnn 1 Show that rank A = n.

 

− −

−

 

140

3.5

CHAPTER 3.

LINEAR TRANSFORMATIONS 

Change of Basis Formulas

In Section 3.3 we learned how to represent a linear transformation as a matrix, and in the last section we saw how the matrix can be used to tell us some facts about the kernel and range of the linear transformation. Clearly the “simpler” the matrix representation the easier it is to understand the linear transformation. Since Gaussian elimination is a method used to obtain a “simple” matrix representation for a system of equations, we have already seen the utility of  finding nice representations. Once we have a fixed linear transformation, that maps a vector space  V   into itself, the only variable in determining its matrix representation is our choice of  basis. Before studying how to pick such a basis (cf. Chapter 5), we need to learn how the matrix representations of the same linear transformation with respect to different bases are related to each other. Thus, let L  be a linear transformation from V   into V , where dim(V ) = n. Let F  = {f j : j = 1, . . . , n} and G = {g j : j = 1, . . . , n} be two bases of V . Let A = [ajk ] be the matrix representation of  L with respect to F  and let B  = [bjk ] be the matrix representation of  L with respect to G. This means that the following equations hold: n

f j ] = L[f 

 

akj f k

 

(3.7)

bkj g k

 

(3.8)

k=1 n

L[gg j ] =

k=1

x]T  x]T  Let P  = [ pjk ] be the change of basis matrix that satisfies [x G  = P [x F , that is n

f j =



 pkj g k

 

(3.9)

k=1

and if  P 

−1

= [q jk ], then

n

g j  =



q kj f k

 

(3.10)

k=1

We refer the reader to Theorem 2.11 and the material preceding it. Computing f j ], we have L[f  n

n

             

f j ] = L L[f 

 pkj g k  =

k=1

n

=

n

n

bskg s  =

 pkj

k=1 n

=

 pkj L[gg k ]

k=1

n

s=1 n

n

n

 pkj bsk

k=1 s=1

(q ms bsk pkj ) f m

m=1

k=1 s=1

m=1

q msf m



141

3.5. CHANGE OF BASIS FORMULAS 

Comparing this expression with (3.7) and equating coefficients, we have n

n



amj  =

q ms bsk pkj

 

(3.11)

k=1 s=1

for m and j  varying from 1 through n. However, the right-hand side of (3.11) is the m, j entry of the matrix  P  1 BP . Thus we have −

−1

A = P 

−1

BP  and B = P AP 

(3.12)

A better way to remember how the matrix representations  A and B  are related is to utilize formulas (3.3), that is, x)]T  x]T  [L(x F  = A[x F 

x)]T  x]T  [L(x G  =  B[x G

Thus, x)]T  x)]T  x]T  [L(x G  = P [L(x F  = P (A[x F ) −1

= (P A)(P 

1 x]T  x]T  [x G ) = (P AP  )[x G −

Hence, we must have B = P AP  1 . The reader is referred to problem 11 in Section 2.6 for a justification of this last step. −

to R2 defined by L(x1 , x2 ) = (2x1 +x2 , 3x1 − 2x2 ). Verify formulas (3.12) for F  =  {(1, 1), (−1, 2)} and G =  { (−1, −1), (2, 0)}. Example 1. Let L   be a linear transformation from

2

R

Solution.   Writing the vectors in F  as linear combinations of the vectors in  G,

we have (1, 1) =  − (−1, −1) + 0(2, 0)

   

(−1, 2) =  − 2(−1, −1) + − Thus,



−1 −2 P  = 0 − 32



−1

P 

=

3 2

(2, 0)

4 −1 3 0 − 32

To determine A, we compute 7  2 (1, 1) − (−1, 2) 3 3 7  7 L(−1, 2) = (0, −7) =  − (1, 1) − (−1, 2) 3 3 L(1, 1) = (3, 1) =

Thus, A =



−

7 3 2 3

− 37 − 37



142

CHAPTER 3.

LINEAR TRANSFORMATIONS 

To determine B , we compute L(−1, −1) = (−3, −1) = (−1, −1) − (2, 0) L(2, 0) = (4, 6) =  − 6(−1, −1) − (2, 0) Thus,







 

1 −6 B = −1 −1 Computing P 

−1



BP  we have

4 −1 3 0 − 23



1 −6 −1 −1

−1 −2  = 0 − 32 −

7 3 2 3



− 37  = A − 37

which is formula (3.12). Example 2. Let L :

 4

R

→ R4 be a linear transformation defined by

L(x1 , x2 , x3 , x4 ) = (2x3  + x4 , −3x1 + x2  − x4 , x1  − x3  + 6x4 , x2  − x3 ) Let S  be the standard basis of R4 and let G =  { (−1, 0, 1, 1), (0, 1, −1, 0), (0, 0, 1, 1), (1, 0, 1, 0)}. Find the matrix representations of  L with respect to S , and then G, by employing (3.12). Let A   be the matrix representation of  L   with respect to S . By inspection we have 0 0 2 1 0 −1 −3 1 A = 1 0 −1 6 0 1 −1 0 Solution.

If  P 

−1

 

 

is the change of basis matrix that satisfies [ x]T  S  = P 

−1

P 

 

0 −1 0 1 = 1 −1 1 0

An easy computation gives us

 

0 0 1 1

−1

[x]T  G  then,

 

1 0 1 0

 

1 1 −1 −1 0 1 0 0 P  = 1 −1 −1 2 0 1 1 −1

Using (3.12), where B is the matrix representation of  L with respect to the basis

143

3.5. CHANGE OF BASIS FORMULAS 

G, we have −1

B = P AP 

   

−1 1 1 −1 0 1 0 0 = 1 −1 −1 2 0 1 1 −1 4 2 = −5 7

 

2 2 −1 1 −1 −3 0 −3 3 0 5 −2

  

0 −3 1 0

0 2 1 0 0 −1 1 −1

  

1 −1 6 0

−1 0 0 1 1 −1 1 0

0 0 1 1

 

1 0 1 0

As a check on our computations we compute  L(−1, 0, 1, 1) using the matrix  B , and then compare this with our original definition of  L. The vector (−1, 0, 1, 1) is the first vector in the basis G. Thus, the first column of  B   contains the coordinates of  L(−1, 0, 1, 1) with respect to the basis  G. Hence, L(−1, 0, 1, 1) = 4(−1, 0, 1, 1) + 2(0, 1, −1, 0) − 5(0, 0, 1, 1) + 7(1, 0, 1, 0) = (3, 2, 4, −1) This is exactly what we get when we compute  L(−1, 0, 1, 1) directly.



The preceding calculations have shown us that if  L : V  → V   is a linear transformation, and A  and  B  are two matrix representations of  L  with respect to two different bases of  V , then there is a matrix  P  such that A  = P  1 BP . −

Definition 3.9. Let A  and  B  be two n × n matrices. We say that  A  is similar −1

to B , if there is a nonsingular matrix  P   such that A  = P 

BP .

As we have seen, the matrix  P  is nothing more than the matrix relating two different bases of our vector space. Moreover, we understand that similar matrices are just different matrix representations of the same linear transformation. The change of basis discussion assumed that the linear transformation L mapped V   into V . What happens when L : V  →  W , and we change bases in V  and W ? Without going through the calculations, we state the appropriate theorem. Theorem 3.8. Let  L : V  → W  be a linear transformation. Let  F  and  F  be two bases of  V  . Let  G  and  G   be two bases of  W . Let  A  be the matrix representation  of  L  using the bases  F  and  G  while  B   is the representation using the bases  F  x]T  x]T  = P [x and  G . Let  P  = [ pjk ]  be the change of basis matrix that satisfies  [x F . F  T  T  y ]G  = Q[yy ]G . Then  Let  Q  = [q jk ]  be the matrix that satisfies  [y

A = Q

−1

BP 

 

(3.13)

The reader should look at problem 9 at the end of this section for one method of  proving this result.

144

CHAPTER 3.

LINEAR TRANSFORMATIONS 

Problem Set 3.5 1. Let L : (2, 5)}.

2

R

→ R2 be defined by L(x1 , x2 ) = (2x1 , 2x2 ). Let F  =  { (1, −1),

a. Find the matrix representation A  of  L  with respect to the standard basis S . x]T  x]T  b. Let P  be the change of basis matrix that satisfies [x F  = P [x S . Find 1 P  and P  . −

c. Find the matrix representation B of  L  with respect to the basis F  by using (3.12). 2. Let A  be any 2 × 2 scalar matrix, that is A =

 

c 0  = cI 2 0 c −1

Let P  be any 2 × 2 nonsingular matrix. Show that P AP  problem 1. 3. Let L :

3

R

equals A; cf.

→ R3 . Let F  = { (1, 2, −3), (1, 0, 0), (0, 1, 0)}. Suppose

 

0 1 A = 2 1 5 0

 

−2 0 1

is the matrix representation of  L  with respect to the basis  F . Use (3.12) to find the matrix representation of  L  with respect to the standard basis; cf. problem 4 of Section 3.2.

 

2 1  is the matrix representation of a linear transfor−1 2 0 1 mation with respect to the standard basis. Let P  =   and set 1 0 B = P AP  1 . Then B can be thought of as the matrix representation of  L with respect to some other basis. What is this basis?

4. Suppose A =

−

 

5. Let A, B, and C  be three n × n matrices. Show that a. A is similar to itself. b. If  A  is similar to B , then B  is similar to A. c. If  A  is similar to B and B  is similar to C , then A  is similar to C . 6. Let A  be any 2 × 2 matrix. Show that if  A  is similar to I 2 , then A = I 2 . 7. Let F  = {(1, 2), (1, 0)} and G = {(1, −1), (0, 1)}. Let L(x1 , x2 ) = (x1  + x2 , 2x1  − x2 ). Find the matrix representations of  L  with respect to bases F  and G. Verify (3.12).

145

3.5. CHANGE OF BASIS FORMULAS 

8. Let F  =  {(1, 0, 1), (0, 1, 1), (1, 1, 0)} and G =  { (1, 0, −1), (−1, 1, 0), (0, 0, 1)}. Let L(x1 , x2 , x3 ) = (x1 + x2 + x3 , x2 + x3 , x3 ). Verify (3.12) for the above bases and linear transformation L. 9. Prove Theorem 3.8. Hint: Use problem 11 in Section 2.6 and formula (3.3). 10. Let F 1  = Let F 2  = Let G 1  =

     

       











0 1 1 0 1 1 1 1 , , , 1 1 1 1 0 1 1 0

1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 , , , , , 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0

0 0 0 0 0 1

Let G 2  =

       

1 0 0 1 0 0 0 0 , , , 0 0 0 0 1 0 0 1



0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 , , , , , 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1

1 1 1 1 1 0

Let  j Ak  be the matrix representation of a linear transformation L : M 22  → M 23  with respect to the bases  G j and F k . That is, x)]T  x]T  [L(x Gj =  j Ak [x F k Suppose that

  

−2 0 −3 1 0 1 0 1 4 1 A1  = 3 1 −5 2 −2 0 1 2 3 Find the other three matrix representations.

0 0 1 0 1 4

  

11. Suppose A and B  are similar n × n matrices, i.e., there is a matrix P   such that A = P BP  1 . −

a. How are ker(A) and ker(B) related? b. How are Rg(A) and Rg(B) related? 12. Let L : P 2  →  P 2  be a linear transformation. Let F  = { t2 +t−1, t2 +2, t−6}. Suppose the matrix representation of  L  with respect to  F  is

 

 

−14 −2 −18 23 11 18 A = 11 2 15

Find the matrix representation of  L  with respect to the standard basis of  P 2 .

146

CHAPTER 3.

LINEAR TRANSFORMATIONS 

Supplementary Problems 1. Define and give examples of each of the following: a. Linear transformation b. Range and kernel of a linear transformation c. Rank of a matrix d. Column (row) rank

2 2. Let A = 1 from

k

R

 

6 −2  be the matrix representation of a linear transformation 3 5 P  to R .

a. k =? p =? b. Determine the dimensions of ker(A) and Rg(A). 3. For each of the following matrices determine the rank and find a basis for the kernel:

 a. 

 

3 1 2 1 0 1 −1 1 1

−1 b.  4

2 3 1 0

0 6 −1 0 0 1

 

4 c. 6

2 4 6 3 6 9 2 1 2 1

 

4. Let A  and  B  be two matrices. Show that Rank(AB) ≤ minimum{rank(A),rank(B)} Hint: Rank (A) equals dim(Rg(A)). 5. Define L : R2 → R2 by L(x1 , x2 ) = (x1  + x2 , −x2 ). Show that L maps the straight line y  = mx onto the straight line y  = −[m/(m +1)]x. What happens to the line y  = −x? 6. Describe, geometrically, the following linear transformations, and in each case determine the kernel and range: a. L(x1 , x2 ) = (x1 − x2 , x1  + x2 ) b. L(x1 , x2 ) = (2x1 − x2 , 4x1 − 2x2 ) 7. Let V 1 , V 2 , and V 3  be vector spaces with bases  F 1 , F 2 , and F 3 , respectively. Suppose L1 : V 1 → V 2 and L 2 : V 2 →  V 3  are linear transformations. Let A1 and A 2  be their corresponding matrix representations with respect to x) =  L2 (L1 (x x)). the given bases. Define L : V 1  → V 3 by L(x a Show that L  is a linear transformation from  V 1 to V 3 . b. Show that the matrix representation of  L with respect to the bases F 1 and F 3 is A 2 A1 .

3.5.

147

CHANGE OF BASIS FORMULAS 

8. A linear transformation L : R2 → R2 is said to be positive if it maps the first quadrant into the first quadrant; that is, if  x1 and x2   are both positive, then so are y1 and y2 , where L(x1 , x2 ) = (y1 , y2 ). a. Let  L (x1 , x2 ) = (ax1 + bx2 , cx1 + dx2 ). Show that  L  is positive if and only if   a,b, c, and d  are all non-negative and at least one of  { a, b}  is positive and one of  { c, d}  is positive. b. Find a positive linear transformation whose matrix representation with respect to some basis has at least one negative entry. c. Find an example of a positive linear transformation whose kernel has dimension equal to 1. 9. A mapping T  from a vector space V    into V    is said to be an affine transformation if  T (x ) =  L (x ) + a where  L  is a linear transformation from  V    into  V    and a is any fixed vector in V   . a. Show that T (x) = L(x) + a  is a linear transformation if and only if  a equals the zero vector. b. Given any straight line in R2 , show that there is an affine transformation that maps the line x2  = 0 onto the given line. Hint: Rotate and then translate. 10. Two vector spaces  V  1 and  V  2  are said to be isomorphic if there is a linear transformation L : V  1  →  V  2  that is both one-to-one and onto. a. Suppose dim(V  1 ) =  m  and V  1  is isomorphic to V  2 . Show dim(V  2 ) = m. b. Show that two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension. c. Let V  1 = R1 and let V  2  be the vector space defined in Example 5 Section 2.2. Define L : V  1 → V  2 by L(x) = ex . Show that L is a linear transformation that is one-to-one and onto. n 11. Let V   = { k=1 ak sin(kx)e−k t , n  any positive integer and ak   arbitrary numbers}. 2

a. Under ordinary addition and multiplication show that V    is a vector u space. Define L : V   →  V   by L[u] = ∂u . − ∂  ∂t ∂x 2

2

b. Show L  is a linear transformation. c. Find the kernel and range of  L.