Download Mathematics 206 Solutions for HWK 23 Section 6.3 p358 §6.3 p358

Mathematics 206
Solutions for HWK 23
Section 6.3 p358
§6.3 p358 Problem 9. Given T (x, y, z) = (13x − 9y + 4z, 6x + 5y − 3z) and v = (1, −2, 1), use
the standard matrix for the linear transformation T to find the image of the vector v.
Solution. Note that the domain for T is R3 and the codomain is R2 . So we’re expecting a
2 × 3 matrix. Since T (1, 0, 0) = (13, 6), T (0, 1, 0) = (−9, 5), and T (0, 0, 1) = (4, −3), the standard
matrix for T is
·
¸
13 −9 4
A=
6
5 −3
and the standard coordinate matrix for T (v) is


·
1
13
A  −2  =
6
1
−9
5

·
¸
1
4 
35
−2  =
.
−3
−7
1
¸

In other words, T (v) = (35, −7).
§6.3 p358 Problem 13. Given that T : R2 −→ R2 is the reflection through the origin, T (x, y) =
(−x, −y), and given v = (3, 4), (a) find the standard matrix A for the linear transformation T , (b)
use A to find the image of the vector v, and (c) sketch the graph of v and its image.
Solution.
(a) Since T (1, 0) = (−1, 0) and T (0, 1) = (0, −1), the standard matrix is
·
¸
0
.
−1
·
0
−1
−1
A=
0
(b)
−1
[T (v)] = A[v] =
0
¸· ¸ ·
¸
3
−3
=
4
−4
(c) See the sketch in the text.
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A. Sontag
May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 15. Given that T : R2 −→ R2 is the counterclockwise rotation of 135◦ in
R2 , and given v = (4, 4), (a) find the standard matrix A for the linear transformation T , (b) use
A to find the image of the vector v, and (c) sketch the graph of v and its image.
Solution.
(a)
1 √ √
(− 2, 2)
2
√
1 √
T (1, 0) = (cos 225◦ , sin 225◦ ) = (− 2, − 2)
2
√ ¸
· √
1 −√ 2 −√2
A=
2 − 2
2
T (1, 0) = (cos 135◦ , sin 135◦ ) =
(b)
· √
1 −√ 2
[T (v)] = A [v] =
2
2
√ ¸· ¸
· √ ¸
−√2
4
− 2
=4
− 2
4
0
√
T (v) = (−4 2, 0)
(c) See the text for a sketch.
§6.3 p358 Problem 17. Given that T : R3 −→ R3 is the reflection through the xy-coordinate
plane in R3 , T (x, y, z) = (x, y, −z), and given that v = (3, 2, 2), (a) find the standard matrix A for
the linear transformation T , (b) use A to find the image of the vector v, and (c) sketch the graph
of v and its image.
Solution.

1
(a) T (1, 0, 0) = (1, 0, 0), T (0, 1, 0) = (0, 1, 0), and T (0, 0, 1) = (0, 0, −1) so A =  0
0

  

1 0 0
3
3
(b) [T (v)] = Av =  0 1 0   2  =  2  so T (v) = (3, 2, −2).
0 0 −1
2
−2

0 0
1 0 .
0 −1
(c) See the text for a sketch.
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A. Sontag
May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 21. Given that T : R2 −→ R2 is the projection onto the vector w = (3, 1)
in R2 , T (v) = projw v, and given that v = (1, 4), (a) find the standard matrix A for the linear
transformation T , (b) use A to find the image of the vector v, and (c) sketch the graph of v and
its image.
Solution. (a)
3
(3, 1)
10
1
T (0, 1) = proj(1,3) (0, 1) =
(3, 1)
10
·
¸
1 9 3
A=
10 3 1
T (1, 0) = proj(1,3) (1, 0) =
(b)
·
1 9
[T (v)] = Av =
10 3
3
1
¸· ¸
· ¸
· ¸
1 21
7 3
1
=
=
4
10 7
10 1
T (v) =
7
(3, 1)
10
(c) See the sketch in the text.
§6.3 p358 Problem 25. Find the standard matrices for T = T2 ◦ T1 and T 0 = T1 ◦ T2 , given
T1 : R2 −→ R2 ,
T1 (x, y) = (x − 2y, 2x + 3y)
T2 : R2 −→ R2 ,
T2 (x, y) = (2x, x − y)
·
¸
1 −2
Solution. The standard matrix for T1 is A1 =
.
2 3
·
¸
2 0
The standard matrix for T2 is A2 =
.
1 −1
In this situation matrix multiplication· corresponds
¸ to composition of functions, so
2 −4
the standard matrix for T is A2 A1 =
−1 · −5
¸
0 2
0
and the standard matrix for T is A1 A2 =
.
7 −3
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A. Sontag
May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 37. Given
T : R2 −→ R3 ,
T (x, y) = (x + y, x, y), v = (5, 4)
B = {(1, −1), (0, 1)}, B 0 = {(1, 1, 0), (0, 1, 1), (1, 0, 1)}
find T (v) by using (a) the standard matrix and (b) the matrix relative to B and B 0 .
Solution.


1 1
(a) The standard matrix for T is A =  1 0  so
0 1

1
[T (v)] = A[v] =  1
0

 
1 · ¸
9
5
0
= 5
4
1
4
and T (v) = (9, 5, 4).
(b) Denote the vectors in B, in the order given, as v1 , v2 . Similarly, let the vectors in B 0 be called
w1 , w2 , w3 . Then T (v1 ) = (0, 1, −1) = w1 − w3 , T (v2 ) = (1, 0, 1) = w3 . Therefore the matrix for
T , relative to B and B 0 is

¯
·
¸
1
¯
A0 = [T (v1 )]B 0 ¯¯ [T (v2 )]B 0 =  0
−1

0
0.
1
Moreover, v = 5v1 + 9v2 . This gives
[T (v)]B 0


 
1 0 · ¸
5
5
= A0 [v]B =  0 0 
= 0
9
−1 1
4
and consequently T (v) = 5w1 + 4w3 = (9, 5, 4), which agrees with the result found in (a).
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May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 41. Given
T : R3 −→ R4 ,
T (x, y, z) = (2x, x + y, y + z, x + z), v = (1, −5, 2)
B = {(2, 0, 1), (0, 2, 1), (1, 2, 1)}, B 0 = {(1, 0, 0, 1), (0, 1, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0)}
find T (v) by using (a) the standard matrix and (b) the matrix relative to B and B 0 .

2
1
Solution. (a) The standard matrix for T is A = 
0
1

2
1
[T (v)] = A[v] = 
0
1
0
1
1
0
0
1
1
0

0
0
 so
1
1

  2 
0 
1
0
 −4 
 −5  = 

1
−3
2
1
3
so T (v) = (2, −4, −3, 3).
(b) Denote the vectors in B by v1 , v2 , v3 , and those in B 0 by w1 , w2 , w3 , w4 . Then
T (v1 ) = (4, 2, 1, 3) = 2w1 + w2 + w3 + w4
T (v2 ) = (0, 2, 3, 1) = −2w1 + 3w2 + 3w3 − w4
T (v3 ) = (2, 3, 3, 2) = −w1 + 3w2 + 3w3 + 0w4
The matrix for T relative to B and B 0 is therefore

2 −2
¯
¯
·
¸
¯
¯
1 3
A0 = [T (v1 )]B 0 ¯¯ [T (v2 )]B 0 ¯¯ [T (v3 )]B 0 = 
1 3
1 −1
Moreover, v = 92 v1 +
11
2 v2

−1
3 

3
0
− 8v3 . Therefore
[T (v)]B 0

2
1
= A0 [v]B = 
1
1
−2
3
3
−1



−1  9 
6
2
3   11
 −3 
 2 =

3
−3
−8
0
−1
which gives T (v) = 6w1 − 3w2 − 3w3 − w4 = (2, −4, −3, 3), which agrees with the result from (a).
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May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 45. Let T : P2 −→ P3 be given by T (p) = xp. (In other words, if p(x) =
c0 + c1 x + c2 x2 , then T (p) is the polynomial defined by (T (p))(x) = x(p(x)) = c0 x + c1 x2 + c2 x3 .
Find the matrix of T relative to the bases B = {1, x, x2 } and B 0 = {1, x, x2 , x3 }.
Solution.
 
0
1
T (1) = x, so [T (1)]B 0 =  .
0
0
 
0
0

T (x) = x2 , so [T (x)]B 0 =  .
1
0
 
0
0

T (x2 ) = x3 , so [T (x2 )]B 0 =  .
0
1
Therefore the required matrix is

0
¯
¯
·
¸
¯
¯
1
[T (1)]B 0 ¯¯[T (x)]B 0 ¯¯[T (x2 )]B 0 = 
0
0
0
0
1
0

0
0
.
0
1
§6.3 p358 Problem 47. Let B = {1, x, ex , xex } be a basis of a subspace W of the space of
continuous functions, and let Dx be the differential operator on W . (In other words Dx : W −→ W
is the linear transformation defined by Dx (f ) = f 0 = the derivative function for the function f .)
Find the matrix for Dx relative to the basis B.
Solution. Dx (1) = 0, Dx (x) = 1, Dx (ex ) = ex , and Dx (xex ) = ex + xex , so the required matrix
is


0 1 0 0
¯
¯
¯
·
¸
¯
¯
¯
0 0 0 0
A = [Dx (1)]B ¯¯ [Dx (x)]B ¯¯ [Dx (ex )]B ¯¯ [Dx (xex )]B
=

0 0 1 1
0 0 0 1
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May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 49. Use the matrix from Exercise 47 to evaluate Dx (3x − 2xex ).
Solution. Let f be the function we wish to differentiate using Exercise 47. Then

0
0

[Dx (3x − 2xex )]B = [Dx (f )]B = A[f ]B = 
0
0
1
0
0
0

 

0
0
3
0 3   0 

=

1
0
−2
1
−2
−2
0
0
1
0
This tells us that Dx (f )(x) = 3 − 2ex − 2xex , exactly as we would expect from calculus.
Let B = {1, x,Z x2 , x3 } be a basis for P3 , and let T : P3 −→ P4 be
x
the linear transformation given by T (xk ) =
tk dt.
§6.3 p358 Problem 51.
0
(a) Find the matrix A for T with respect to B and the standard basis for P4 .
(b) Use A to integrate p(x) = 6 − 2x + 3x3 .
Solution. (a)
0
T (1) = T (x ) =
Z
x
1 dt = x,
T (x) =
0
T (x2 ) =
Z
x
t2 dt =
0
3
x
,
3
T (x3 ) =
Z
Z
x
t dt =
0
x
t3 dt =
0
Give the name B 0 to the standard basis for P4 : B 0 = {1, x, x2 , x3 , x4 }.
for T is

0
¯
¯
¯
·
¸ 1
¯
¯
¯

A = [ T (1)]B 0 ¯¯ [T (x)]B 0 ¯¯ [T (x2 )]B 0 ¯¯[T (x3 )]B 0 =  0

0
0
x2
2
x4
4
Then the required matrix
0
0
1
2
0
0
0
0
0
1
3
0

0
0

0

0
1
4
(b) The instructions are a little vague. Let’s assume that what’s wanted is to find T (6 − 2x + 3x3 ).
Then we have
[T (6 − 2x + 3x3 )]B 0

0
1

£
¤

= A 6 − 2x + 3x2 B =  0

0
0
0
0
1
2
0
0
0
0
0
1
3
0



0 
0

6
0
 6 
  −2  

0
 =  −1  .
 0


0
0
3
1
3
4
4
Thus T (6 − 2x + 3x3 ) = 6x − x2 + 34 x4 .
Page 7 of 8
A. Sontag
May 2, 2001
Math 206 HWK 23 Solns contd
6.3 p358
§6.3 p358 Problem 55. Let T : M2,3 −→ M3,2 be given by T (A) = AT . Find the matrix for T
relative to the standard bases for M2,3 and M3,2 .
Solution. Having consulted the text on p201, and following the order suggested by Example 5,
I’ll take the standard basis for M2,3 to be {M1 , M2 , M3 , M4 , M5 , M6 }, where M1 , M2 , and M3 have
zeros in all positions of the bottom row and all except one position of the top row and they have
1 in the first, second, third positions, respectively of the first row. Then M4 , M5 , M6 have zeros in
all positions of the top row and two positions of the bottom row, with 1 in the first, second, and
third positions, respectively of the second row. (Yes, I’m trying to avoid having to type in all those
matrices.) I’ll take the standard basis for M3,2 , which I’ll write as B 0 = {N1 , N2 , N3 , N4 , N5 , N6 },
to be arranged in like fashion: first let the 1’s move across the first row, then across the second
row, and finally across the third row, always from left to right. If you ordered these two bases
differently, your representing matrix will come out different from mine, but your results should be
consistent with mine once you take that difference into account. So here goes, finally.
T (M1 ) = (M1 )T = N1 ,
T (M2 ) = (M2 )T = N3
T (M3 ) = (M3 )T = N5 ,
T (M4 ) = (M4 )T = N2
T (M5 ) = (M5 )T = N4 ,
T (M6 ) = (M6 )T = N6
Therefore the matrix we want is

1
0

0

0

0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
1
0
0
Page 8 of 8

0
0

0
.
0

0
1
A. Sontag
May 2, 2001