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Math 110 Homework 11
Partial Solutions
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5.2.12 (a) Let
be the eigenspace of T corresponding to , and let −1
be the eigenspace of T−1 corresponding to −1 . If x ∈ ≥ then
,
−1
−1
−1
T(x ) = x . Thus x = T (x ) = T
(x ) and so T (x ) = −1 .
Thus x ∈ −1 . By the same argument, we also get −1 ⊆ ≥
and thus that they are equal.
(b) If T is diagonalizable, then there exists a basis ﬁ for such that
[T]ﬁ is diagonal. By (a), [T−1 ]ﬁ must then also be diagonal.
5.2.20 Suppose that
= W1 ⊕ · · · ⊕
. Then, by Theorem 5.10, we can
ﬁnd ordered bases ﬁ1
k for W1
k , respectively, so that ﬁ =
ﬁ1 ∪ · · · ∪ is a basis for . This then gives that dim( ) = dim(W1 ) +
· · · + dimWk by counting basis vectors.
On the other hand, suppose that the dimension equality holds. Picking
a basis ﬁi for each Wi, we have that ﬁ = ﬁ1 ∪· · ·∪ has dim( ) vectors.
Since W1 + · · · + Wk = , we also have that ﬁ spans . Thus ﬁ is a
basis for , and so = W1 ⊕ · · · ⊕
.
5.4.13 Suppose that W is the Tcyclic subspace of
generated by v and
that w ∈ . Since W = span({
(v) T 2 (v)
}), this means that
there exists an integer n and a0 1
∈
such that w = a0 +
n(v) = (a + a + · · · + a Tn)(v). Setting g(t) =
a1 (v) + · · · + anT
0
1
n
a0 + a1 + · · · + antn, we have that w = g(T)(v).
5.4.17 Let f (t) = ± + cn−1 −1 + c 1 + c0 . By the CayleyHamilton Theorem, f ( ) = 0. In other words, n = ±(cn−1 −1 + · · · + c1 + c0 ).
−1
+k
Thus n ∈ span({
}). Inductively, if n n+1
∈
−1
n+k+1 = ±(c
+k
k+2
span({
}), then
+ · · · + c1
+
n−1
−1
2
c + 0 k+1 ) ∈ span({
}). Thus span({
}) =
−1
span({
}) has dimension ≤ .
5.4.18 (a) Note that det( ) = det( − 0I ) = f (0) = a0 . Thus
if and only if det( ) = 0 if and only if a0 = 0. is invertible 2
(b) By the CayleyHamilton Theorem, we have (−1)n An + an−1 An−1 +
a1 A + a0 I = 0. This can be rearranged to give −a0 I = (−1)n An +
an−1 An−1 + a1 A. Since A is invertible (and thus by part (a),
a0 = 0), we can multiply this entire equation by −a−1 A−1 to give
0
the desired result.
5.4.20 If there exists a g such that U = g (T ), then it is clear that U T = T U .
On the other hand, suppose that U T = T U . Suppose that V is the T cyclic subspace of itself generated by v . By exercise 5.4.13 (using U (v )
in place of w), we can ﬁnd a polynomial g so that g (T )(v ) = U (v ). We
then claim that U = g (T ). Notice ﬁrst that U (T k (v )) = T k (U (v )) =
T k (g (T )(v )) = g (T )(T k (v )) for every k ≥ 0. Then, if x ∈ V is any
vector, we can write x as some linear combination of ﬁnitely many of
the vectors T k (v ). By linearity, we see (calculation suppressed) that
U (x) = g (T )(x). Since this is true for any x ∈ V , we have that U =
g (T ). ...
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 Spring '10
 FUCKHEAD
 Linear Algebra, WI, 0 K, 2 span, 1 Span

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