Partial Solution Set, Leon §4.3 4.3.2 Let

Partial Solution Set, Leon §4.3
4.3.2 Let [u1, u2] and [v1, v2] be ordered bases for R 2 , where u1 = (1, 1) T , u2 = (−1, 1) T ,
v1 = (2, 1) T , and v2 = (1, 0) T . Let L be the linear transformation defined by L(x) =
(−x1, x2) T , and let B be the matrix representing L with respect to [u1, u2].
(a) Find the transition matrix S corresponding to the change of basis from [u1, u2] to
[v1, v2].
Solution: This part doesn’t deal with L yet, rather just the change of basis matrix.
The transition matrix in question is the one I’ve been calling TUV , i.e.,
S = V −1 U =
� 0 1
1 −2
� � 1 −1
1 1
� �
=
1 1
−1 −3
(b) Find the matrix A representing L with respect to [v1, v2] by computing A = SBS −1 .
Solution: Note that S goes from the [u1, u2] basis to [v1, v2] basis, which is not
the way the diagrams are shown in the book.
Now, B is the matrix that corresponds to the [u1, u2] is obtained by
B = U −1
� �
L(e1)L(e2) U = 1
� � � � � � � �
1 1 −1 0 1 −1 0 1
=
2 −1 1 0 1 1 1 1 0
�
.
U

Next we find S−1 = 1
� �
3 1
. Then it is a simple matter to determine that
2 −1 −1
A = SBS −1 � � � � � � � �
1 1 0 1 1 3 1 1 0
=
=
.
−1 −3 1 0 2 −1 −1 −4 −1
V
1. Verify that L(v1) = a11v1 + a21v2 and L(v2) = a12v1 + a22v2
� � � � � � � �
2 −2
Solution: Note that L v1 = L = and on the other hand
� � � �
1
� �
1
2 1 −2
a11v1 + a21v2 = 1 · + (−4) = , so they match.
1 0 1
� � � � � � � �
1 −1
Also L v2 = L = and on the other hand
�
0
�
0
� � � �
2 1 −1
a12v1 + a22v2 = 0 · + (−1) = , so they match.
1 0 0
4.3.3 Let L be the linear transformation on R 3 given by
L(x) = (2x1 − x2 − x3, 2x2 − x1 − x3, 2x3 − x1 − x2) T ,
and let A be the matrix representing L with respect to the standard basis for R 3 . If
u1 = (1, 1, 0) T , u2 = (1, 0, 1) T , and u3 = (0, 1, 1) T , then [u1, u2, u3] is an ordered basis
for R 3 .
(a) Find the transition matrix U corresponding to the change of basis from [u1, u2, u3]
to the standard basis.
(b) Determine the matrix B representing L with respect to [u1, u2, u3].
Solution:
⎡
(a) This is simply U = ⎣
1 1 0
1 0 1
0 1 1
⎤
⎦ .
2

(b) Somewhat surprisingly, B = U −1 AU = A. An interesting sidelight: this means that
UA = AU, i.e., we have an instance of a commuting pair of matrices.
4.3.4 Let L be the linear operator mapping R3 into R3 ⎡
⎤
defined by L(x) = Ax, where A =
3 −1 −2
⎣ 2 0 −2 ⎦ . Let v1 = (1, 1, 1)
2 −1 −1
T , v2 = (1, 2, 0) T , and v3 = (0, −2, 1) T . Find the
transition matrix V corresponding to a change of basis from [v1, v2, v3] to the standard
basis, and use it to determine the matrix B representing L with respect to [v1, v2, v3].
⎡ ⎤
1 1 0
Solution: The transition matrix is V = ⎣ 1 2 −2 ⎦ . We want
1 0 1
B = V −1 ⎡
−2 1
⎤ ⎡
2 3 −1
⎤ ⎡
−2 1 1
⎤
0
⎡
0 0
⎤
0
AV = ⎣ 3 −1 −2 ⎦ ⎣ 2 0 −2 ⎦ ⎣ 1 2 −2 ⎦ = ⎣ 0 1 0 ⎦ .
2 −1 −1 2 −1 −1 1 0 1 0 0 1
4.3.5 Let L be the linear operator on P3 defined by
L(p(x)) = xp ′ (x) + p ′′ (x).
(a) Find the matrix A representing L with respect to [1, x, x 2 ].
(b) Find the matrix B representing L with respect to basis B given by [1, x, 1 + x 2 ].
(c) Find the matrix S such that B = S −1 AS.
(d) Given p(x) = a0 + a1x + a2(1 + x 2 ), find L n (p(x)).
(e) Given p(x) = a0 + a1x + a2(x 2 ), find L n (p(x)) (this is not in the book, but try it!).
Solution:
(a) We start by applying L to the basis vectors: L(1) = 0, L(x) = x, and L(x2 ) = 2x2 ⎡ ⎤+
0 0 2
2. The corresponding coordinate vectors become the columns of A = ⎣ 0 1 0 ⎦ .
0 0 2
(b) The coordinate vectors for 1 and x are unchanged, but the coordinate vector for
2x2 + 2 is now (0, 0, 2) T ⎡ ⎤
0 0 0
, so B = ⎣ 0 1 0 ⎦ .
0 0 2
(c) The change of basis matrix has for its columns the vectors in R3 corresponding
to each of the vectors in [1, x, 1 + x2 ] since we have to find the change of basis
matrix from the arbitrary basis [1, x, 1 + x2 ] to the standard basis [1, x, x2 ⎡ ⎤
]: S =
1 0 1
⎣ 0 1 0 ⎦ .
0 0 1
3

(d) The coordinate vector of p(x) is (a0, a1, a2) T B
with respect to the basis B given by
[1, x, 1 + x2 ]. The nth power of L is given by the nth power of B, which is simple
to compute because of the simple diagonal structure of B; Bn ⎡
0 0 0
= ⎣ 0 1 0
0 0 2n ⎤
⎦ . It
follows that the coordinate vector for L n (p(x)) is B n (a0, a1, a2) T = (0, a1, 2 n a2), so
L n (p(x)) = a1x + 2 n a2(1 + x 2 ).
(e) There isn’t a part (e), but I would like to emphasis how this B = S−1AS helps:
The coordinate vector of p(x) is (a0, a1, a2) T with respect to standard basis. The
nth power of L is given by the nth power of A, which is not so simple to computer
since A is not diagonal as it was in part (d). However, from B = S−1AS we can
find A = SBS −1 , so then A2 = (SBS −1 ) 2 = SBS −1SBS −1 = SB2S −1 . Similarly,
A100 = SB100S−1 , or generally An = SBnS −1 , for all natural numbers n. Thus
⎡
A n = ⎣
1 0 1
0 1 0
0 0 1
⎤ ⎡
⎦ ⎣
0 0 0
0 1 0
0 0 2 n
⎤ ⎡
⎦ ⎣
1 0 1
0 1 0
0 0 1
which is easy to computer as a product of three matrices, instead of multiplying A
by itself n times. It follows that the coordinate vector for L n (p(x)) is A n (a0, a1, a2) T
which is
4.3.8 Suppose that A = SΛS −1 , where Λ is a diagonal matrix with main diagonal λ1, λ2, . . . , λn.
(a) Show that Asi = λisi for each 1 ≤ i ≤ n.
n�
(b) Show that if x = αisi, then Ak n�
x =
i=1
i=1
αiλ k i si.
(c) Suppose that |λi| < 1 for each 1 ≤ i ≤ n. What happens to A k x as k → ∞?
Solution:
(a) For any choice of i, 1 ≤ i ≤ n, we have
Asi = � SΛS −1� =
si
SΛ � S −1 �
si
= SΛei
= S (Λei)
4
= Sλiei
= λiSei
= λisi.
⎤
⎦
−1
,

The reason that S −1 si = ei is because when you multiply S −1 S = I, and so when
you multiply the ith column of S by S−1 you get the ith column of I, which is ei.
(b) This is easily proven by induction: A0 n�
x = x = αisi.
Assume that A k x =
n�
i=1
and the result follows by induction.
i=1
αiλ k i si for some k ∈ N. Then
A k+1 x = AA k x
= A
=
=
=
=
n�
i=1
n�
i=1
n�
i=1
n�
i=1
n�
i=1
αiλ k i si
Aαiλ k i si
αiλ k i Asi
αiλ k i λisi
αiλ k+1
i si,
(c) Each term in the preceding sum vanishes, since if |λ| < 1 then lim
k→∞ λ k = 0.
4.3.9 Suppose that A = ST , where S is nonsingular. Let B = T S. Show that B is similar to
A.
Proof: Assume that A is as described, i.e., that A = ST and that S is nonsingular.
Then
B = T S = (S −1 S)T S = S −1 (ST )S = S −1 AS,
so B is similar to A.
A second way to present the proof above would be:
Proof: Assume that A is as described, i.e., that A = ST and since S is nonsingular
T = S −1 A. Then
B = T S = (S −1 A)S = S −1 AS,
so B is similar to A.
What’s the point? Given any square S and T , with at least one of the two nonsingular,
we know that it’s unlikely that ST = T S. But at least ST and T S are similar. And that
(as we shall see) means that they have much in common (eigenvalues, for example).
5

4.3.11 Show that if A and B are similar matrices, then det(A) = det(B).
Solution: If A and B are similar, then B = S −1 AS for some nonsingular S. Now use
the fact that the determinant of a product is the product of the determinants, along with
the commutativity of real multiplication: det B = det(S −1 AS) = det S −1 det A det S =
det S −1 det S det A = det(S −1 S) det A = det I det A = det A.
6