Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 5
John K. Nguyen
December 7, 2011
5.2.11. Let A be an n n matrix that is similar to an upper triangular matrix and has the distinct eigenvalues 1 , 2 , ..., k with corresponding multiplicities m1 , m2 , ..., mk . Prove the following statements.
(a) tra(A) =

k
X

mi i .

i=1

(b) det(A) = (1 )m1 (2 )m2 ...(k )mk .

Proof of (b). Since A be an n n matrix that is similar to an upper triangular matrix, say B, that has the
distinct eigenvalues 1 , 2 , ..., k with corresponding multiplicities m1 , m2 , ..., mk . By definition of similar,
there exists an invertible matrix Q such that A = Q1 BQ. Recall Theorem 4.7 which states that for
any A, B M22 (F ), det(AB) = det(A)det(B). In consideration of this theorem and the corollary on
1
page 223, we have that det(A) = det(Q1 BQ) = det(Q1 )det(B)det(Q) = det(Q)
det(B)det(Q) = det(B) =
m1
m2
mk
(1 ) (2 ) ...(k )
as required.
5.2.12. Let T be an invertible linear operator on a finite-dimensional vector space V .
(a) Recall that for any eigenvalue of T , 1 is an eigenvalue of T 1 . Prove that the eigenspace of T
corresponding to is the same as the eigenspace of T 1 corresponding to 1 .
(b) Prove that if T is diagonalizable, then T 1 is diagonalizable.
Proof of (a). Pick v E . Then, by definition, T (v) = v. Taking the inverse of both sides, we get
T 1 T (v) = T 1 (v) which means v = T 1 (v). Then, by definition, v E1 so we have that the
eigenspace of T corresponding to is the same as the eigenspace of T 1 corresponding to 1 .
Proof of (b). Suppose T is diagonalizable. Then T has n linearly independent eigenvectors. From part (a)
we know that T 1 also has the same n eigenvectors. Thus, T 1 is diagonalizable.
5.2.13. Let A Mnn (F ). Recall from Exercise 14 of Section 5.1 that A and At have the same characteristic polynomial and hence share the same eigenvalues with the same multiplicities. For any eigenvalue of
A and At , let E and E0 denote the corresponding eigenspaces for A and At , respectively.
(a) Show by way of example that for a given common eigenvalue, these two eigenspaces need not be the
same.
(b) Prove that for any eigenvalue , dim(E ) = dim(E0 ).
(c) Prove that if A is diagonalizable, then At is also diagonalizable.


Example for (a). Define A =

1
3

2
4

. Now, notice that A(1, 0) = (1, 3) but At (1, 0) = (1, 2). Thus,

t
EA 6= EA
.

Proof of (b). Suppose dim(F ) = n. Then, by definition, we know that dim(E ) = n rank(A I). Then,
taking the transpose (recall that rank(A) = rank(At ) by a previous exercise), we have that dim(E ) =
n rank((A I)t ) = n rank(At I) = dim(E0 ) as required.
Proof of (c). Suppose A is diagonalizable. Then, there exists an invertible matrix Q such that B = Q1 AQ
is a diagonal matrix (recall that a square matrix is diagonalizable if it is similar to a diagonal matrix according
to Section 5.1 ). Now, taking the transpose of both sides yields B t = (Q1 AQ)t = Qt At (Q1 )t . Clear,y B t
is diagonal so by definition, we have that At is diagonalizable.
5.2.18a. Prove that if T and U are simultaneously diagonalizable operators, then T and U commute (i.e.,
U T = T U ).
Proof. Suppose T and U are simultaneously diagonalizable operators. Then, by definition, there exists an
ordered basis = {v1 , v2 , ..., vn } such that T (vi ) = vi and U (vi ) = vi where i = 1, 2, ..., n. It follows that
T U (vi ) = T (U (vi )) = T (vi ) = T (vi ) = vi = vi = U (vi ) = U (vi ) = U (T (vi )) = U T (vi ). So, we
can conclude that U T = T U as required.
5.4.13. Let T be a linear operator on a vector space V , let v be a nonzero vector in V , and let W be the
T-cyclic subspace of V generated by v. For any w V , prove that w W if and only if there exists a
polynomial g(t) such that w = g(T )(v).
Proof. () Suppose w W and assume dim(W ) = n. Let = {v, T v, ..., T n1 v} be an ordered basis for
W . Then, by definition, w = a0 v + a1 T v + ... + an1 T n1 v for some scalars a0 , a1 , ...an1 (that is, w is a
linear combination of the elements of ). Let g(t) = a0 + a1 t + ... + an1 tn1 . Then, g(t) is a polynomial of
t of degree less than or equal to n 1. This also means that w = g(T )v.
() Suppose there exists a polynomial g(t) such that w = g(T )v. Then, g(t) = a0 + a1 t + ... + an tn
and w = a0 v + a1 T v + ... + an T n v for some scalars a0 , a1 , ..., an . Since v, T v, ..., T n v W , w W .
5.4.16. Let T be a linear operator on a finite-dimensional vector space V .
(a) Prove that if the characteristic polynomial of T splits, then so does the characteristic polynomial of
the restriction of T to any T-invariant subspace of V .
(b) Deduce that if the characteristic polynomial of T splits, then any nontrivial T-invariant subspace of V
contains an eigenvector of T .
Proof of (a). Suppose the characteristic polynomial of T , say f , splits. Let W be a T invariant subspace
of V and g is the characteristic polynomial of TW . Then, g divides f so there exists a polynomial r such
that f = gr. Suppose that f has n degrees. Note that the amount of zeros that g have is less than or equal
to degree of g and analogously for r. But, it follows that n = deg(g) = deg(f ). So, g has deg(g) zeros, h has
deg(h) zeros. Thus, we can factor g to deg(g) factors which means that the characteristic polynomial of the
restriction of T to a T-invariant subspace splits.
Proof of (b). Suppose W is a nontrivial T-invariant subspace of V . Let f be the characteristic polynomial
of TW . By part (a), we know that f splits. Pick to be the root of f . Then, f () = det(TW I) = 0 .
But this means that TW I is not invertible so there exists a nonzero w W such that (T I)(w) = 0
which implies that w is an eigenvector of T . Since w is arbitrary, we have shown that if the characteristic
polynomial of T splits, then any nontrivial T-invariant subspace of V contains an eigenvector of T .
5.4.20. Let T be a linear operator on a vector space V , and suppose that V is a T-cyclic subspace of itself.
Prove that if U is a linear operator on V , then U T = T U if and only if U = g(T ) for some polynomial g(t).
Hint: Suppose that V is generated by v. Choose g(t) according to Exercise 13 so that g(T )(v) = U (v).

Proof. () Suppose U T = T U . Then, since V = span(v, T (v), T 2 (v), ...), U (v) = a0 + a1 T (v) + ... + an T n (v)
for some scalars a0 , a1 , ..., an . So, U (v) = g(T )(v) where g(v) = a0 + a1 x + ... + an xn . Suppose x V . Then,
x = b0 + b1 T (v) + ... + bm T m (v) for some scalars b0 , b1 , ..., bm . It follows that
U (x) = U (b0 + b1 T (v) + ... + bm T m (v)) = b0 U (T 0 (v)) + b1 U (T (v)) + ... + bm U (T m (v))
= b0 T 0 (U (v)) + b1 T (U (v)) + ... + bm T m (U (v))
= b0 T 0 (g(T )(v)) + b1 T (g(T )(v)) + ... + bm T m (g(T )(v))
= b0 g(T )(T 0 (v)) + b1 g(T )(T (v)) + ... + bm g(T )(T m (v))
= g(T )(b0 T 0 (v) + b1 T 1 (v) + ... + bm T m (v))
= g(T )(x).
Thus, U = g(T ) for some polynomial g.
() Suppose U = g(T ) for some polynomial g. Define g(x) = a0 + a1 x + ... + an xn . Then,
U T (x) = a0 T 0 (T (x)) + a1 T (T (x)) + ... + an T n (T (x))
= a0 T (T 0 (x)) + a1 T (T (x)) + ... + an T (T n (x))
= T (a0 T 0 (x) + a1 T (x) + ... + an T n (x))
= T U (x).
Therefore, we have that U T = T U .
We have shown that if U is a linear operator on V , then U T = T U if and only if U = g(T ) for some
polynomial g(t).