MATH 205 HOMEWORK #4 OFFICIAL SOLUTION
Problem 1: Suppose that (v1 , . . . , vn ) is a basis of V such that the matrix of T with respect to
that basis is in Jordan form. Describe the matrix of T with respect to the basis (vn , . . . , v1 ).
Solution: Since T is a matrix in Jordan form, we know that T vi = λi vi + i vi−1 , where λi is the
i-th entry on the diagonal of the matrix of T , and i is either 1 or 0. Let wi = vn−i ; we want to
describe the matrix of T with respect to {w1 , . . . , wn }. We have
T wi = T vn−i = λn−i vn−i + n−i vn−i−1 = λn−i wi + n−i wi+1 .
Thus the diagonal of T with respect to {w1 , . . . , wn } has diagonal entries {λn , . . . , λ1 } and zeroes
elsewhere except below the diagonal, where it has the entries {n−1 , . . . , 1 }.
Problem 2: Give an example of a vector space V of dimension at least 2 over an algebraically
closed field F and a linear transformation V → V such that V does not decompose into two
nontrivial invariant subspaces.
Solution: Let V = F 2 , and let T be represented by
3 1
.
0 3
Then if V split into two invariant subspaces U and W then V would have two eigenvectors. However,
V only has one: 3, and only one eigenvector, (1, 0). Thus V does not decompose into two nontrivial
invariant subspaces.
Problem 3: For this problem, F = R and V = Rn . Give an example of an invertible linear
transformation which has only zeroes on its diagonal. Give an example of a non-invertible linear
transformation which has only nonzero entries on the diagonal.
Solution: Let {e1 , . . . , en } be the standard basis for Rn . Let T map ei to ei−1 , with e0 = en ; this
has 1’s above the main diagonal, and a 1 in the lower-right corner. Thus it has 0’s on the diagonal
but is invertible.
Now let T be the transformation which takes ei to e1 + · · · + en , for all i. This is non-injective
and thus non-invertible, but its matrix consists entirely of 1s.
Problem 4:
(a) Suppose that p(x) = ax2 + bx + c, Show that if u ∈ ker p(T ) then span(u, T u) is an invariant
subspace of T .
(b) Suppose that V is a finite dimensional vector space over R. Show that any linear transformation T : V → V has a one or two-dimensional invariant subspace. (Hint: any real
polynomial factors into a product of linear and quadratic polynomials.)
Solution:
(a) In order to show that U = span(u, T u) is an invariant subspace it suffices to show that
T 2 u ∈ U , since
T span(u, T u) = span(T u, T 2 u).
But since u ∈ ker p(T ) we know that
aT 2 u + bT u + cu = 0.
1
2
MATH 205 HOMEWORK #4 OFFICIAL SOLUTION
If a 6= 0 then T 2 u = −b/aT u − c/au ∈ span(u, T u). If a = 0 then u satisfies
bT u + cu = 0.
If b 6= 0 then T u = −c/bu, so T 2 u = (c2 /b2 )u ∈ span(u, T u). If b = 0 then cu = 0, so u = 0
(in which case span(u, T u) = 0).
(b) Write dim V = n. Then we know that the vectors v, T v, . . . , T n v are linearly dependent, so
there exist coefficients a0 , . . . , an , not all zero, such that
an T n v + · · · + a0 v = (an T n + · · · + a0 I)v = 0.
Write p(x) = an xn + · · · + a1 x + a0 ; we can factor
m
Y
p(x) =
(ai x2 + bi x + ci ).
i=1
Then
m
Y
(ai T 2 + bi T + ci I)v = 0;
i=1
so for some i, the transformation (ai T 2 + bi T + ci I) has a nontrivial kernel. Let u be in this
kernel; then by part (a), span(u, T u) is an invariant subspace of T of dimension at most 2,
as desired.
Problem 5: Suppose that S is another linear transformation such that S has n distinct eigenvalues
and such that all eigenvectors of S are also eigenvectors of T . Show that ST = T S.
Solution: Since S has n distinct eigenvalues there is a basis v1 , . . . , vn of V such that Svi = λi vi
for each i. Since each eigenvector of S is an eigenvector of T we know that there exist ρi ∈ F such
that T vi = ρi vi for all i. Then we know that
n
n
n
n
n
X
X
X
X
X
ST
ai vi = S
ai ρi vi =
ai ρi λi vi = T
ai λi vi = T S
ai vi .
i=1
i=1
i=1
i=1
i=1
Since every vector in V is in the span of the vi , ST v = T Sv for all v ∈ V , and thus ST = T S.
Problem 6: The goal of this problem is to show that if n is odd and F = R then T must have
an eigenvalue.
(a) Suppose that we can write V = U ⊕ W . Let PW,U be the function that takes a vector v,
written v = u + w with u ∈ U and w ∈ W , to w. Show that PW,U is a well-defined linear
transformation. Prove that PW,U + PU,W is the identity transformation on V .
(b) For any w ∈ W , show that PW,U ◦ T is a well-defined linear transformation on W .
(c) Suppose that U is an invariant subspace of T . Show that if λ is an eigenvalue of PW,U ◦ T
with eigenvector w ∈ W then it is an eigenvalue of T . (Hint: consider a vector of the form
u + w with u ∈ U and apply T − λI to it. Where does it land?)
(d) Prove that if n is odd and F = R then T has an eigenvalue.
Solution:
(a) Since V = U ⊕ W every vector v can be written uniquely as u + w with u ∈ U and w ∈ W .
In particular, if v = u + w and v 0 = u0 + w0 then
v + v 0 = (u + u0 ) + (w + w0 ),
so PW,U (v + v 0 ) = w + w0 = PW,U v + PW,U v 0 . The proof that PW,U (λv) = λPW,U v works
analogously.
Let v = u + w. Then
(PW,U + PU,W )v = PW,U v + PU,W v = w + u = v.
MATH 205 HOMEWORK #4 OFFICIAL SOLUTION
3
Thus PW,U + PU,W is the identity transformation on V .
(b) We know that T : W → V is a linear transformation, and that PW,U : V → W . Thus
PW,U ◦ T is a linear transformation W → W .
(c) Note that
(T − λI)(u + aw) = (T − λI)u + (T − λI)aw = T u − λu + PU,W T aw + (PW,U T aw − λaw) ∈ U.
|
{z
}
=0
Thus (T − λI)(span(U, w)) ⊆ U , which is a subspace of lower dimension. Thus dim ker(T −
λI) ≥ 1, and λ is an eigenvalue of T .
(d) If n = 1 then T clearly has an eigenvalue. We will prove the general case by induction.
Suppose that dim V = 2k + 1. From problem 4 we know that T has an invariant subspace
U with dimension at most 2. If dim U = 1 then we’re done: any nonzero vector in U is an
eigenvector, and T has an eigenvalue. If dim U = 2, let W be such that V = U ⊕ W . Then
dim W = 2k − 1, and by the inductive hypothesis W has an eigenvalue λ. From part (c) we
know that λ is also an eigenvalue of T , and we’re done.
Problem 7: For this problem, V does not have to be finite dimensional. Suppose that there exist
distinct λ1 , . . . , λm ∈ F such that
(T − λ1 I) · · · (T − λm I) = 0.
Let Ui = ker(T − λi I), the λi -eigenspace. Show that
V = U1 ⊕ · · · ⊕ Um .
(Hint: if you have trouble, try the case when m = 2, λ1 = 0 and λ2 = 1.)
Solution: We know that eigenvectors with different eigenvalues are linearly independent. Thus we
know that U = U1 ⊕ · · · ⊕ Um ⊆ V ; we just need to show that it spans all of V . Let W be such that
V = U ⊕W , and let w ∈ W . If w 6= 0 then there exists some k such that (T −λk I) · · · (T −λn I)w ∈ U
but (T − λk+1 I) · · · (T − λn I)w is not. Thus we can write (T − λk+1 I) · · · (T − λn I)w = w0 + u.
Then
(T − λk I)(w0 + u) = (T − λk I)w0 + (T − λk I)u ∈ U,
so (T − λk I)w0 ∈ U . Write
(T − λk I)w = u0 = u1 + · · · + uk−1 + uk+1 + · · · + un
with ui ∈ Ui . (We can write it this way because u0 ∈ ker(T −λ1 I) · · · (T −λk−1 I)(T −λk+1 I) · · · (T −
λn I).) Consider the vector
1
1
1
1
u1 − · · · −
uk−1 −
uk+1 − · · · −
un .
v = w0 −
λ1 − λk
λk−1 − λk
λk+1 − λk
λn − λk
Then
(T − λk )v = (T − λk I)w0 − u1 − · · · − un = 0,
0
so v ∈ Uk and thus w ∈ U , a contradiction. The only way this is possible is if w = 0, and thus W
is trivial.
Problem 8: Let F = R, n = 2 and suppose that T is represented by the matrix
−3 −4
2
3
in terms of the standard basis. Explain what T k looks like for all k.
Solution: Note that T 2 = I. Thus T odd = T and T even = I.