Problem Set 7 Solutions
MATH 110: Linear Algebra
Each problem is worth 5 points.
PART 1
1. Curtis p. 192 3.
2. Curtis p. 192 5.
3. Curtis p. 215 1(a,b,c,d,e).
Solutions in book.
PART 2
Problem 1(10)
a) If T:V→Vhas an eigenvalue λ, prove that aT has the eigenvalue
aλ.
Proof: Let xbe an eigenvector for the eigenvalue λ. (aT )(x) = a(T(x)) =
a(λx) = λ(ax) so ax is an eigenvector of aT with eigenvalue λ.
b) If xis an eigenvector for both T1and T2, prove that xis an eigenvector
for aT1+bT2and find the eigenvalues of aT1+bT2in terms of the eigenvalues
of T1and T2.
Proof: Let λbe the eigenvalue of xfor T1and let µbe the eigenvalue of
xfor T2. (aT1+bT2)(x) = (aT1)(x)+(bT2)(x) = (λa +µb)x.
c) Suppose that xis an eigenvector of Twith eigenvalue λ. Show that x
is an eigenvector of T2with eigenvalue λ2.
Proof:T2(x) = T(T(x)) = T(λx) = λ(T(x)) = λ(λx) = λ2x.
d) Let Pbe a polynomial. Show that if xis an eigenvector of Twith
eigenvalue λthen xis an eigenvector of P(T) with eigenvalue P(λ).
Proof: By induction on the degree of P(we will denote the degree with
the variable n). Suppose n= 0: the polynomial is a constant, and certainly
xis an eigenvector of P(T) (every vector actually is) with eigenvalue P(λ).
Now suppose the theorem is true for all polynomials of degree less than or
equal to n. Consider Pof degree n+ 1. If Phas no constant term, then
P(T) = T Q(T) where Qhas degree n. By induction xis an eigenvector of
Qwith eigenvalue Q(λ). Therefore, T Q(T)(x) = T(Q(λ)x) = Q(λ)T(x) =
Q(λ)(λx) = λQ(λ)x. Thus xis an eigenvector of Pwith eigenvalue λQ(λ) =
P(λ). Now, if Phas a constant term c, then P−cis a polynomial of degree
1
