Math 110 Homework 14
May 23, 2014
1 8.A.3
Suppose T L(V ), m is a positive integer, and v V is such that T
m1
v = 0 but T
m
v = 0.
Prove that
(v, Tv, T
2
v, . . . , T
m1
v)
is linearly independent.
Proof : Let a
1
, a
2
, . . . , a
m
F such that a
1
v + a
2
Tv + . . . + a
m
T
m1
v = 0. Then by acting
T
m1
on the equation, we have:
T
m1
(a
1
v + a
2
Tv + . . . + a
m
T
m1
v) = T
m1
(0)
a
1
T
m1
v + a
2
T
m
v + . . . + a
m
T
2(m1)
v = 0
But since T
m
v = 0 and null(T
m
) null(T
m+k
) for any k 0, this becomes
a
1
T
m1
v + 0 + 0 + . . . + 0 = 0
a
1
T
m1
v = 0
Now since T
m1
v = 0, this implies a
1
= 0. This process can be repeated for each coecient a
j
for
j = 2, 3, . . . , m by acting T
mj
on the linear combination, and so we nd a
1
= a
2
= . . . = a
m
= 0.
Therefore the list (v, Tv, T
2
v, . . . , T
m1
v) is linearly independent.
2 8.A.4
Suppose T L(C
3
) is dened by T(z
1
, z
2
, z
3
) = (z
2
, z
3
, 0). Prove that T has no square root.
More precisely, prove that there does not exist S L(C
3
) such that S
2
= T.
Proof : By inspection, null(T) = span(z
1
), so dim null(T) = 1. If there exists S L(C
3
) such
that S
2
= T, then dim null(S) < 1, since dim null(T) = dim null(T
2
) = 2. Therefore dim
null(S) = 0, but then S is injective and surjective, so S
2
= T.
1
3 8.A.5
Give an example of a linear operator T on a vector space such that T
4
= T
3
but T
3
= T
2
.
Let T be the left-shift operator described in Exercise 8.A.4, i.e. T L(C
3
) dened by
T(z
1
, z
2
, z
3
) = (z
2
, z
3
, 0). Then T
2
(z
1
, z
2
, z
3
) = (z
3
, 0, 0) and T
3
(z
1
, z
2
, z
3
) = (0, 0, 0), so T
3
=
T
2
. But T
4
(z
1
, z
2
, z
3
) = T(T
3
(z
1
, z
2
, z
3
)) = T(0, 0, 0) = (0, 0, 0), so T
4
= T
3
.
4 8.A.8
Suppose that S, T L(V ). Prove that if ST is nilpotent, then TS is nilpotent.
Proof : If ST is nilpotent, then there exists a positive integer m such that (ST)
m
= 0, or
(ST)(ST) . . . (ST) = 0
S(TS)(TS) . . . (TS)T = S(TS)
m1
T = 0
Then left composing the equation with T yields
TS(TS)
m1
T = T(0) = 0
And right composing with S yields
TS(TS)
m1
TS = 0S = 0
(TS)
m+1
= 0
Therefore TS is nilpotent.
5 8.A.12
Suppose V is an inner product space. Prove that if N L(V ) is self-adjoint and nilpotent, then
N = 0.
Proof : By 8.17, if N is a nilpotent operator on V , then there exists a basis of V such that
M(N) is upper diagonal with all of its diagonal elements equal to zero. Then M(N
) is a lower
diagonal matrix with all of its diagonal elements equal to zero.
If N is self-adjoint, then M(N) = M(N
). This implies that each element of M(N) is equal to
zero, so N = 0.
6 8.A.13
Suppose V is an inner product space. Prove that if N L(V ) is normal and nilpotent, then
N = 0.
Proof : Since N is a nilpotent operator, all its eigenvalues must be zero. Since N is a normal
operator, M(N) is diagonalizable so N = UDU
1
where D is a diagonal matrix with the
eigenvalues of N along its diagonal (i.e. D = 0.) Therefore N = U0U
1
= 0.
2
7 8.B.2
Give an example of an operator T on a nite-dimensional real vector space such that 0 is the
only eigenvalue of T, but T is not nilpotent.
Let T be an operator such that
M(T) =
0 0 0
0 0 1
0 1 0
Then T has only one real eigenvalue, which is zero, but T is not nilpotent.
8 8.B.3
Suppose F = C and 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 = ND.
Proof : Since V is a vector space over C, then there exists a basis (v
1
, v
2
, . . . , v
n
) of V with
respect to which T is upper triangular. Thus there is a portion of M(T) which is diagonalizable,
so let M(D) be the matrix, of dimension equal to dim(M(T)), containing the diagonalizable
portion of M(T) and zeros elsewhere. Then the operator N = T D will have a matrix that
is upper triangular with zeros on the diagonal, which denes a nilpotent operator. Thus, by
construction, T = D + N.
9 8.B.4
Suppose that V is an n-dimensional complex vector space and T is a linear operator on V such
that
null(T
n2
) = null(T
n1
)
Prove that T has at most two distinct eigenvalues.
Proof : If T is nilpotent, then its only eigenvalue is zero and the claim is true, so suppose T
is not nilpotent. Then by the ascending containment chain of nullspaces for a linear operator,
null(T
n2
) = null(T
n1
) implies dim null(T
n1
) n 1. Also, since T is not nilpotent by
assumption, dim null(T
n
) n1. Therefore null(T
n1
) = null(T
n
), so any non-zero eigenvalue
of T must also be an eigenvalue of V/null(T
n1
), which is a one-dimensional space. Thus T has
at most two eigenvalues.
10 8.B.5
Suppose V is a 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 V has a basis consisting of eigenvectors of T, then M(T) is diagonal with
respect to that basis. If dim V = n, then we can say that
1
, . . . ,
m
are the distinct eigenvalues
of T, where m n. Then we can write a basis for V as (v
1,1
, . . . , v
1,i
, . . . , v
m,1
, . . . , v
m,i
) where
3
each v
k
is an eigenvector of T with eigenvalue
k
. This implies (v
j,1
, . . . , v
j,1
) is a basis for
null(T
j
I), thus
V = null(T
1
I) null(T
m
I)
Then since we also have V = null(T
1
I)
n
null(T
m
I)
n
, this implies null(T
j
I) =
null(T
j
I)
n
. Therefore each generalized eigenvector of T is also an eigenvector of T.
Now suppose each generalized eigenvector is also an eigenvector of T, and let
1
, . . . ,
m
be
the eigenvalues of T. Let v null(T
j
I)
n
. Then v is a generalized eigenvector of T with
eigenvalue
j
, so by assumption v is also an eigenvector of T with eigenvalue
j
. Therefore
null(T
j
I)
n
= null(T
j
I), so by concatenating bases for each of the m nullspaces, we have
a basis consisting of eigenvectors of T.
11 8.B.6
Prove or give a counterexample: If V is a complex vector space and dim V = n and T L(V ),
then T
n
is diagonalizable.
Let T L(C
2
) be an operator such that
M(T) =
1 1
0 1
Then the matrix for T
2
is given by
M(T
2
) =
1 1
0 1
1 1
0 1
1 2
0 1
This matrix is not diagonalizable, so the claim is false.
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