Math 344: Linear Algebra
Homework 24, Solutions
8.A.1 Define T ∈ L(C2 ) by
T (w, z) = (z, 0).
Find all generalized eigenvectors of T .
Solution: The matrix of T in the standard basis is given by
0 1
M (T ) =
,
0 0
therefore the only eigenvalue of T is 0. Also note that M (T )2 =
0, so C2 = G(0, T ). This means that every vector of C2 is a
generalized eigenvector of T .
8.A.3 Suppose that T ∈ L(V ) is invertible. Prove that G(λ, T ) =
G( λ1 , T −1 ) for every λ ∈ F with λ 6= 0.
Proof. Fix λ ∈ F with λ 6= 0. Suppose that v ∈ G(λ, T ).
Then there is a positive integer n such that (T − λI)n v = 0.
Multiplying this equation on the left by (−λ−1 T −1 )n we have
0 = (−λ−1 T −1 )n (T − λI)n v
= (−λ−1 T −1 (T − λI))n v
= (T −1 − λ−1 I)n v.
This shows that v ∈ G(λ−1 , T −1 ). We conclude that
G(λ, T ) ⊆ G(λ−1 , T −1 ).
The same argument with the roles of λ and T being played by
λ−1 and T −1 shows that
G(λ−1 , T −1 ) ⊆ G(λ, T ).
8.A.5 Suppose that T ∈ L(V ), m is a positive integer, and v ∈ V is
such that T m−1 v 6= 0 but T m v = 0. Prove that
v, T v, T 2 v, . . . , T m−1 v
is linearly independent.
Proof. We start by noting that the hypotheses imply that v 6= 0.
Suppose that a0 , . . . , am−1 are scalars in F such that
a0 v + · · · + am−1 T m−1 v = 0.
1
2
Apply the operator T m−1 to this equation to see that a0 T m−1 v =
0; since T m−1 v 6= 0, we have a0 = 0. Therefore
a1 T v + · · · + am−1 T m−1 v = 0.
Next apply the operator T m−2 to this last equation to see that
a1 T m−1 v = 0; then a1 = 0. Continuing in this manner we
will conclude that all of the scalars a0 , . . . , am−1 are equal to 0.
Hence
v, T v, T 2 v, . . . , T m−1 v
is linearly independent.
8.A.7 Suppose that N ∈ L(V ) is nilpotent. Prove that 0 is the only
eigenvalue of N .
Proof. Let λ ∈ F be an eigenvalue of N . Suppose that N n = 0
for some nonnegative integer n. Let v ∈ V be an eigenvector of
N corresponding to λ. Then N v = λv, so
0 = N n v = λn v;
and so λn = 0. We conclude that λ = 0.
8.A.8 Prove or give a counterexample: the set of nilpotent operators
on V is a subspace of L(V ).
This statement is false.
Counterexample: Consider the operators S, T ∈ L(C2 ) defined by
S(w, z) = (0, w);
and
T (w, z) = (z, 0).
2
2
Then S = T = 0, so S and T are nilpotent. Also,
(S + T )(w, z) = (z, w),
2
so (S + T ) = I, which implies that S + T is not nilpotent.
We have shown that the set of nilpotent operators on a vector
space V need not be closed under addition.
8.B.1 Suppose that V is a finite-dimensional complex vector space,
N ∈ L(V ) and that 0 is the only eigenvalue of N . Prove that
N is nilpotent.
Proof. By 8.21 (a) and 8.11 we have
V = G(0, N ) = null (N dim V ).
Therefore N dim V is the zero operator on V . We conclude that
N is nilpotent.
3
8.B.2 Give an example of an operator T on a finite-dimensional real
vector space such that 0 is the only eigenvalue of T but T is
not nilpotent.
Example: Let T ∈ L(R3 ) be defined by
T (x, y, z) = (−y, x, 0).
Note that T (0, 0, 1) = (0, 0, 0) so 0 is an eigenvalue of T . Now
suppose that λ 6= 0 and T (x, y, z) = λ(x, y, z). It follows that
z = 0, and λx = −y, λy = x. Therefore (λ2 +1)x = 0, so x = 0,
which then implies that y = 0. We conclude that 0 is the only
eigenvalue of T . Finally it is straightforward to compute
T 4 (x, y, z) = (x, y, 0)
for all (x, y, z) ∈ R3 , so we can see that T is not nilpotent.
8.B.3 Suppose that T ∈ L(V ). Suppose that S ∈ L(V ) is invertible.
Prove that T and S −1 T S have the same eigenvalues with the
same multiplicity.
Proof. Let k be a nonnegative integer. Then
(S −1 T S)k = (S −1 T S)(S −1 T S) · · · (S −1 T S) = S −1 T k S;
also for any operators T1 and T2 ,
S −1 (T1 + T2 )S = S −1 T1 S + S −1 T2 S.
It follows for any polynomial p ∈ P(F), that
p(S −1 T S) = S −1 p(T )S.
Let n = dim V , and fix some λ ∈ F. Applying the last
statement to the polynomial p(z) = (z − λ)n , we have
(S −1 T S − λI)n = S −1 (T − λI)n S.
We now claim that S −1 affords a bijection
G(λ, T ) → G(λ, S −1 T S).
Note that the truth of this claim implies that T and S −1 T S
have the same eigenvalues with the same multiplicity. Let v ∈
G(λ, T ). Then
[S −1 (T − λI)n S](S −1 v) = S −1 (T − λI)n v = 0,
which proves that S −1 maps G(λ, T ) to G(λ, S −1 T S). Similarly,
for w ∈ G(λ, S −1 T S),
(S −1 T S − λI)n w = 0 =⇒ S −1 (T − λI)n Sw = 0
=⇒ (T − λI)n Sw = 0,
4
which shows that Sw ∈ G(λ, T ). Then since w = S −1 (Sw) we
conclude that the map
S −1 : G(λ, T ) → G(λ, S −1 T S)
is bijective.
8.B.5 Suppose that V is a finite-dimensional complex vector space and
T ∈ L(V ). Prove that V has a basis consisting of eigenvectors
of T if and only if every generalized eigenvector of T is an
eigenvector of T .
Proof. Suppose that V has a basis consisting of eigenvectors of
T . Let λ1 , . . . , λm be the distinct eigenvalues of T . Then
V = E(λ1 , T ) ⊕ · · · ⊕ E(λm , T ).
Moreover, in any case,
V = G(λ1 , T ) ⊕ · · · ⊕ G(λm , T ).
Since E(λi , T ) ⊆ G(λi , T ) for i ∈ {1, . . . , m}, we conclude that
G(λi , T ) = E(λi , T );
this means that every generalized eigenvector of T is an eigenvector of T .
Now assume that every generalized eigenvector of T is an
eigenvector of T . Then
G(λi , T ) = E(λi , T ),
for all i ∈ {1, . . . , m}. This then implies, using 8.23, that
V = E(λ1 , T ) ⊕ · · · ⊕ E(λm , T ).
Hence V has a basis consisting of eigenvectors of T .
8.B.10 Suppose that V is a finite-dimensional complex vector space and
that T ∈ L(V ). Prove that there exist D, N ∈ L(V ) such that
T = D + N , the operator D is diagonalizable, N is nilpotent,
and DN = N D.
Proof. We begin by writing
V = G(λ1 , T ) ⊕ · · · ⊕ G(λm , T ),
where λ1 , . . . , λm are the distinct eigenvalues of T . We define
an operator D ∈ L(V ) by setting
D|G(λi ,T ) = λi IG(λi ,T ) ,
5
where IG(λi ,T ) denotes the identity operator on G(λi , T ). Clearly
the matrix of D with respect to any basis coming from the given
decomposition of V is a diagonal matrix, so D is diagonalizable.
Now set N = T − D. Note that
N |G(λi ,T ) = (T − λi I)|G(λi ,T ) ,
and since the last operator is nilpotent, we conclude that N is
nilpotent. Since we have T = D +N , the proof will be complete
once we prove that DN = N D. To see this note that
DN = D(T − D) = DT − D2 ,
and
N D = (T − D)D = T D − D2 .
Then since D is a scalar operator on G(λi , T ), we see that
DT = T D when restricted to G(λi , T ), but then the fact that
V decomposes as a direct sum of the G(λi , T ) implies that
DT = T D. We conclude that DN = N D.