MATH 5718, ASSIGNMENT 7 – DUE: 19 MAR 2015
LUKE NELSEN
[5B1] Suppose T ∈ L(V ) and there exists a positive integer such that T n = 0.
(a) Prove that I − T is invertible and that (I − T )−1 = I + T + T 2 + · · · + T n−1 .
(b) Explain how you would guess the formula above.
Proof.
(a) Let p be the polynomial defined by p(x) = 1 − x and let q be the polynomial defined by q(x) =
1 + x + x2 + · · · + xn−1 . Then we have
(I − T )(I + T + T 2 + · · · + T n−1 ) = p(T )q(T ) = pq(T ) = q(T )p(T ) = (I + T + T 2 + · · · + T n−1 )(1 − T )
and also
pq(x) = 1 − xn ⇒ pq(T ) = I − T n = I.
So I − T has an inverse, which means that I − T is invertible.
(b) Without previously knowing the inverse, one may guess it by thinking of I − T as 1 − x and wanting
1
, which may be done using the Taylor expansion 1 + x + x2 + x3 + · · · .
to find the operator form of 1−x
Generally, this may not work, but becauce T n = 0 the series terminates and we can define an operator
from 1 + x + x2 + · · · + xn−1 .
[5B2] Suppose T ∈ L(V ) and (T − 2I)(T − 3I)(T − 4I) = 0. Suppose λ is an eigenvalue of T . Prove that
λ = 2 or λ = 3 or λ = 4.
Proof. Let (λ, v) be an eigencouple of T . By Exercise 5B10 (previous homework), we have
0 = (T − 2I)(T − 3I)(T − 4I)v = (λ − 2)(λ − 3)(λ − 4).
Thus λ = 2, 3, or 4.
[8A5] Suppose 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.
Pm−1
Proof. Let a0 , a1 , . . . , am−1 ∈ F such that i=0 ai T i v = 0. Then
! m−1
m−1
X
X
m−1
i
0=T
ai T v =
ai T i+m−1 v = a0 T m−1 v.
i=0
Since T
m−1
v 6= 0, this means a0 = 0. So
0 = T m−2
i=0
Pm−1
m−1
X
i
ai T v = 0 and we have
! m−1
X
i
ai T v =
ai T i+m−1 v = a1 T m−1 v.
i=1
i=1
m−1
Pm−1
i=1
i
Since T
v 6= 0, we have a1 = 0. So i=2 ai T v = 0 and applying T m−3 yields a2 = 0. Proceeding in
this manner, we obtain a0 = a1 = · · · = am−1 = 0. Thus v, T v, T 2 v, . . . , T m−1 v is linearly independent. 1
[8A7] Suppose N ∈ L(V ) is nilpotent. Prove that 0 is the only eigenvalue of N .
Proof. By Proposition 8.19, N has an upper triangular matrix with respect to some basis of V in which all
entries on the diagonal are 0s. By Proposition 5.32, 0 is the only eigenvalue of N .
[8A8] Prove or give a counterexample: The set of nilpotent operators on V is a subspace of L(V ).
Proof. We provide a counterexample: Let T1 , T2 ∈ L(C2 ) be defined by T1 (z1 , z2 ) = (0, z1 ) and T2 (z1 , z2 ) =
(z2 , 0). Since T12 = 0 = T22 , T1 and T2 are nilpotent operators on C2 . However, T1 + T2 is the operator that
maps (z1 , z2 ) 7→ (z2 , z1 ). So (T1 + T2 )2 = I and is certainly not nilpotent. So the set of nilpotent operators
is not closed under addition and is not a vector space.
[8A9] Suppose S, T ∈ L(V ) and ST is nilpotent. Prove that T S is nilpotent.
Proof. Since ST is nilpotent, there is some nonnegative integer m such that (ST )m = 0. Thus
(T S)m+1 = T (ST )m S = 0. So T S is nilpotent.
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[8A16] Suppose T ∈ L(V ). Show that V = range T 0 ⊃ range T 1 ⊃ range T 2 ⊃ · · · .
Proof. Let k ∈ N and let v ∈ range T k+1 . Then there is w ∈ V such that T k+1 w = v. So we have
T k (T w) = v and thus v ∈ range T k . Hence range T k+1 ⊂ range T k for all k ∈ N. Observing that
range T ⊂ V = range IV = range T 0 completes the proof.
[8A17] Suppose T ∈ L(V ) and that m is a nonnegative integer such that range T m = range T m+1 . Prove
that range T k = range T m for all k > m.
Proof. We proceed by induction on k, using k = m + 1 as the base case. So let k ≥ m + 2 and assume that
range T k−1 = range T m . So we have
range T k−1 = range T m
T (range T k−1 ) = T (range T m )
range T k = range T m+1 = range T m .
[8A18] Suppose T ∈ L(V ). Let n = dim V . Prove that range T n = range T n+1 = range T n+1 = · · · .
Proof. If range T k = range T k+1 for some k < n, then we are done by Exercise 8A17. So suppose that
V = range T 0 ) range T 1 ) · · · ) range T n , which means that dim(range T k ) ≤ dim(range T k−1 ) − 1
for all k = 1, 2, . . . , n. Combining these n inequalities yields dim(range T n ) ≤ dim(range T 0 ) − n = 0.
Since range T n+1 ⊂ range T n , we have dim(range T n+1 ) = dim(range T n ) = 0 and therefore we have
range T n+1 = range T n .
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[8A19] Suppose T ∈ L(V ) and m is a nonnegative integer. Prove that null T m = null T m+1 if and only
if range T m = range T m+1 .
Proof. First note that the statement is not necessarily true in either direction if V is infinite-dimensional:
• First, consider V = C∞ and consider T ∈ L(C∞ ) defined by T (z1 , z2 , . . . ) = (0, z1 , z2 , . . . ). So
observe that null T 0 = null T = {0} and range T 0 6= range T .
• Second, consider V = C∞ and consider T ∈ L(C∞ ) defined by T (z1 , z2 , . . . ) = (z2 , z3 , . . . ). So
observe that null T 0 6= null T and range T 0 = range T = V .
So we assume that dim V = n. Thus by Proposition 8.5, V = null T n ⊕ range T n and thus we have
n = dim(null T n ) + dim(range T n ). By the Fundamental Theorem of Linear Maps, we also have that
n = dim(null T m ) + dim(range T m ).
(⇒) Suppose null T m = null T m+1 . By Proposition 8.3, null T m = null T n and thus dim(null T m ) =
dim(null T n ). Thus from our above equalities, dim(range T m ) = dim(range T n ). But from Exercise 8A16
we have range T m ⊃ range T n , and so range T m = range T n .
(⇐) Suppose range T m = range T m+1 . By Exercise 8A17, range T m = range T n and thus dim(range T m ) =
dim(range T n ). Thus from our above equalities, dim(null T m ) = dim(null T n ). But from Proposition 8.2
we have null T m ⊂ null T n , and so null T m = null T n .
[8B1] Suppose V is a complex vector space, N ∈ L(V ), and 0 is the only eigenvalue of N . Prove that N
is nilpotent.
Proof. Since this chapter assumes that V is finite-dimensional (say of dimension n), by Proposition 5.27 we
have that N has an upper-triangular matrix M(N ) with respect to some basis of V . Then since 0 is the
only eigenvalue of N , by Proposition 5.32 the entries on the diagonal are all 0. So each power of M(N )
forces another subdiagonal above the diagonal to have all entries be 0; hence M(N )n is the n × n 0-matrix.
Therefore N n = 0 and N is nilpotent.
[8B2] 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.
Proof. Consider T ∈ L(R3 ) defined by T (x1 , x2 , x3 ) = (0, −x3 , x2 ). Since T (1, 0, 0) = 0 · (1, 0, 0), we have
that 0 is an eigenvalue of T . Now suppose that there is some eigencouple (λ, v) with λ 6= 0 and with v
written as (v1 , v2 , v3 ). So (0, −v3 , v2 ) = T v = (λv1 , λv2 , λv3 ). So 0 = λv1 , which implies v1 = 0. Also,
applying the substition v2 = λv3 (from the third components) to the second components yields −v3 = λ2 v3 ,
which requires v3 = 0 or λ2 = −1. Since λ ∈ R, we have v3 = 0 and thus v2 = 0. But this means v = 0,
which contradicts that v is an eigenvector. So in fact, λ = 0 is the only eigenvalue of T .
Now, T 2 (x1 , x2 , x3 ) = (0, −x2 , x3 ), so T 4 (0, x2 , x3 ) and thus T 4k (0, 1, 0) = (0, 1, 0) 6= 0 for all k ∈ N. So
T cannot be nilpotent.
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