MATH 5718, ASSIGNMENT 6 – DUE: 17 MAR 2015
LUKE NELSEN
[5A18] Show that the operator T ∈ L(C∞ ) defined by T (z1 , z2 , . . . ) = (0, z1 , z2 , . . . ) has no eigenvalues.
Proof. Suppose there exists an eigencouple (λ, z) for T . Write z = (z1 , z2 , . . . ). If λ = 0, then 0 = λz =
T z = (0, z1 , z2 , . . . ) and thus zi = 0 for all i ∈ N. But then z = 0 and z is not an eigenvector. So suppose
λ 6= 0; then (λz1 , λz2 , . . . ) = λz = T z = (0, z1 , z2 , . . . ). So λz1 = 0 and λzi = zi−1 for each i ≥ 2. Since
λ 6= 0, this means that z1 = 0 and by induction we have zi = 0 for all i ∈ N. So z = 0 and z is not an
eigenvector. So in fact there can be no eigencouple λ, z).
[5A21] Suppose T ∈ L(V ) is invertible.
(a) Suppose λ ∈ F with λ 6= 0. Prove that λ is an eigenvalue of T if an only if
(b) Prove that T and T −1 have the same eigenvectors.
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λ
is an eigenvalue of T −1 .
Proof. (a)
λ is an eigenvalue of T ⇔ There exists a nonzero v ∈ V such that T v = λv
⇔ There exists a nonzero v ∈ V such that T −1 T v = T −1 λv
⇔ There exists a nonzero v ∈ V such that v = λT −1 v
1
⇔ There exists a nonzero v ∈ V such that v = T −1 v
λ
1
⇔ is an eigenvalue of T −1 .
λ
Proof. (b) Note that since T is invertible, T is injective. Suppose (0, v) is an eigencouple of T . Then T v = 0;
thus v = 0 and in fact there cannot be any eigenvectors of T corresponding to 0.
Now, suppose v is an eigenvector of T corresponding to λ 6= 0. By the proof of (a), v is an eigenvector of
T −1 corresponding to λ1 . So all eigenvectors of T are eigenvectors of T −1 ; reversing the roles of T and T −1
yields the reverse inclusion and completes the proof.
[5A23] Suppose V is finite-dimensional and S, T ∈ L(V ). Prove that ST and T S have the same eigenvalues.
Proof. Let (λ, v) be an eigencouple of ST . Suppose first that T v 6= 0. Then we have T S(T v) = T (ST v) =
T (λv) = λ(T v), which means that λ is an eigenvalue of T S.
Now suppose T v = 0. Then ST v = 0, which implies that λ = 0. If v ∈ range S, then there is a nonzero
w ∈ V such that Sw = v since v is nonzero and we have T Sw = T v = 0 = λw. Is v ∈
/ range S, then S is not
surjective and thus S is not injective (since V is finite-dimensional). Then there is a nonzero w ∈ V such
that Sw = 0; thus T Sw = T 0 = 0 = λw. In either case, λ = 0 is also an eigenvalue of T S.
So all eigenvalues of ST are eigenvalues of T S. Reversing the roles of S and T yields the reverse inclusion
and completes the proof.
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[5B4] Suppose P ∈ L(V ) and P 2 = P . Prove that V = null P ⊕ range P .
Proof. Let v ∈ V . Then
Pv = PPv
Pv − PPv = 0
P (v − P v) = 0
and thus v − P v ∈ null P . Since P v ∈ range P , we have v = (v − P v) + P v ∈ null P + range P . So
V = null P + range P .
Now let v ∈ null P ∩ range P . Since v ∈ range P , there exists some w ∈ V such that P w = v. Then
v = Pw = PPw = Pv = 0
since v ∈ null P . So null P ∩ range P = {0} and therefore V = null P ⊕ range P .
[5B10] Suppose T ∈ L(V ) and v is an eigenvector of T with eigenvalue λ. Suppose p ∈ P(F). Prove that
p(T )v = p(λ)v.
Proof. We have that T v = λv. First note that for any p ∈ N, we have
T p v = T p−1 (T v) = T p−1 (λv) = λT p−1 v
= λT p−2 (T v) = λT p−2 (λv) = λ2 T p−2 v
..
.
= λp−1 T v = λp−1 (λv) = λp v.
Pn
Then since p ∈ P(F), we have that p(x) = i=1 ai xi for some n ∈ N and a1 , . . . an ∈ F. So
!
n
X
i
p(T )v =
ai T v
i=1
=
n
X
ai (T i v)
i=1
=
=
n
X
ai (λi v)
i=1
n
X
!
i
ai λ
v
i=1
= p(λ)v
[5B13] Suppose W is a complex vector space and that T ∈ L(W ) has no eigenvalues. Prove that every
subspace of W invariant under T is either {0} or infinite-dimensional.
Proof. Suppose U 6= {0} is a finite-dimenional subspace of W that is invariant under T . Then consider
T |U ∈ L(U ): by Proposition 5.21, T |U has an eigenvalue. So let (λ, v) be an eigencouple for T |U ; then
T |U v = λv and thus T v = λv. Then (λ, v) is an eigencouple for T , which contradicts our assumption.
Hence any subspace of W invariant under T cannot be nonzero and finite-dimensional; it must be {0} or
infinite-dimensional.
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[5C1] Suppose T ∈ L(V ) is diagonalizable. Prove that V = null T ⊕ range T .
Proof. Since T is diagonalizable, M(T ) exists for some basis of V . By the definition of a matrix of a linear
map (Definition 3.32), the basis of V is of finite length and thus V is finite-dimensional.
Now, suppose that T is injective. Since V is finite-dimensional, this implies that T is surjective and thus
we are done since V = range T = {0} ⊕ range T = null T ⊕ range T .
Now suppose that T is not injective. So there exists a nonzero v ∈ V such that T v = 0 = 0v. Thus v is
an eigenvector of T corresponding to 0. So 0 is an eigenvalue
Lmof T . Now, since V is finite dimensional and T
is diagonalizable, by Proposition 5.41 we have that V = i=1 Eλi , where λ1 , . . . , λm are theL
eigenvalues of
m
T . Without loss of generality, let λ1 = 0. Then
E
=
null
(T
)
and
we
have
V
=
null
T
⊕
(
i=2 Eλi ). So
Lm λ1
all that remains to show is that range T = i=2 Eλi .
Let v ∈ range T . Since by Proposition 5.41 we know that there is a basis v1 , . . . , vm of V with each
vi anP
eigenvector corresponding to λi . So there is some w P
∈ V such that T
Pmcan write
Pwm = v, and we
m
m
=
w = i=1 αi vi for some α1 , . . . , αm ∈ F. Then v = T w =L i=1 αi (T vi ) = i=1 αi λi vL
i
i=2 αi λi vi .
m
m
Since αi λi vi ∈ EL
for
each
i
=
2,
.
.
.
,
m
we
have
that
v
∈
E
.
Hence
range
T
⊂
E
λi
i=2 λi
i=2 λi and we
m
have range T = i=2 Eλi .
[5C2] Prove the converse of the statement of 5C1 or give a counterexample to the converse.
Proof. We provide a counterexample. Let V = C∞ and define T = IV . Then we have
V = range T = {0} ⊕ range T = null T ⊕ range T.
However, T cannot be diagonalizable since there is no matrix of T .
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