MATH 5718, ASSIGNMENT 5 – DUE: 3 MAR 2015
LUKE NELSEN
[3D16] Suppose V is finite-dimensional and T ∈ L(V ). Prove that T is a scalar multiple of the identity
if and only if ST = T S for every S ∈ L(V ).
Proof. (⇒) Suppose T = λIV for some λ ∈ F. Then for any s ∈ L(V ) we have
ST = S(λIV ) = λ(SIV ) = λ(IV S) = (λIV )S = T S.X
(⇐) Suppose that for some v ∈ V there is no α ∈ F such that T v = αv. So T v, v are linearly independent
in V . Since V is finite dimensional (say of dimension m), we may extend T v, v to a basis v, T v, v1 , ..., vm−2
for V . But then consider the linear operator S defined by mapping the basis elements as follows:
S(T v) = v
Sv = v
Svi = 0 for all i = 1, .., m − 2.
Then S(T v) = Sv = v 6= T v = T (Sv), which contradicts our hypothesis. So in fact, for all v ∈ V there
exists αv ∈ F such that T v = αv v.
Now, since V is finite dimensional there is some basis v1 , ..., vm . Let α := αv1 +···+vm . Then T (v1 + · · · +
vm ) = α(v1 +· · ·+vm ) = αv1 +· · ·+αvm , and also T (v1 +· · ·+vm ) = T v1 +· · ·+T vm = αv1 v1 +· · ·+αvm vm .
So (α − αv1 )v1 + · · · + (α − αvm )vm = 0. Since v1 , ..., vm is linearly independent, we have α = αvi for each
i = 1, ..., m. Hence T vi = αvi for each i = 1, ..., m.
So let v ∈ V . Then v = β1 v1 + · · · + βm vm for some unique β1 , ..., βm ∈ F. So
T v = T (β1 v1 + · · · + βm vm ) = β1 T v1 + · · · + βm T vm = β1 αv1 + · · · + βm αvm = α(β1 v1 + · · · + βm vm ) = αv
and since v ∈ V was arbitrary we have T = αIV . X
[3F33] Let m, n ∈ N. Prove that the function that takes A to At is a linear map from Fm,n to Fn,m .
Furthermore, prove that this linear map is invertible.
Proof. First we see that ·t is a function since each A ∈ Fm,n has only one transpose.
Second we observe that [(αA + βB)t ]ji = (αA + βB)ij = α(A)ij + β(B)ij = α(At )ji + β(B t )ji for each
j = 1, ..., m and each i = 1, ..., n. Thus (αA + βB)t = αAt + βB t and therefore ·t is a linear transformation.
Third, note that At = 0n×m implies (At )ji = 0 for each j = 1, ..., n and each i = 1, ..., m, which implies
(A)ij = 0 for each i = 1, ..., m and each j = 1, ..., n, which implies A = 0m×n . Hence ·t is injective. Also
note that for any B ∈ Fn,m , we have that B t ∈ Fm,n and that (B t )t = B. So ·t is surjective. Therefore ·t is
invertible.
1
[4.6] Suppose p ∈ P(C) has degree m ≥ 0. Prove that p has m distinct zeros if and only if p and its
derivative p0 have no zeros in common.
Proof. Suppose p has k distinct roots z1 , ..., zk where each zi has multiplicity mi . So we write p(z) =
Pk
αΠki=1 (z − zi )mi , where α ∈ C. Note that p0 (z) = α i=1 (z − zi )mi −1 Πj6=i (z − zj )mi .
Pk
(⇒) Suppose m1 = m2 = · · · = mm = 1. Then p0 (z) = α i=1 Πj6=i (z − zj ). So for each zi , we have
p0 (zi ) = αΠj6=i (zi − zj ) 6= 0.
(⇐) Suppose m` ≥ 2 for some ` ≤ k. Then
p0 (z` ) =α
k
X
(z` − zi )mi −1 Πj6=i (z` − zj )mi
i=1
=α
X
=α
X
(z` − zi )mi −1 Πj6=i (z` − zj )mi
i6=`
(z` − zi )mi −1 · 0
i6=`
=0.
[4.9] Suppose p ∈ P(C). define q : C → C by q(z) = p(z)p(z). Prove that q is a polynomial with real
coefficients.
Pm
Pm
Pm
Pm
Proof. Let p(z) = i=0 αi z i , where
αi = ai + bi i. Then p(z) = i=0 αi z i = i=0 αi z i = i=0 αi z i . So
Pm Pi
i
i
p(z)p(z) = i=1
j=0 αj αi−j z . So the coefficient of z of p(z)p(z) is
i
X
αj αi−j =
j=0
i
X
(aj + bj i)(ai−j + bi−j i)
j=0
=
i
X
(aj + bj i)(ai−j − bi−j i)
j=0
=
i
X
(aj ai−j + bj bi−j ) + (ai−j bj − aj bi−j )i)
j=0




i
i
X
X
=  (aj ai−j + bj bi−j ) + i  (ai−j bj − aj bi−j )
j=0

=
i
X
(aj ai−j
j=0



i
i
X
X
+ bj bi−j ) + i 
ai−j bj −
aj bi−j 
j=0
j=0
j=0




i
i
i
X
X
X
=  (aj ai−j + bj bi−j ) + i 
ai−j bj −
ai−j bj 
j=0
=
j=0
j=0
i
X
(aj ai−j + bj bi−j ),
j=0
which is real.
2
[3F8(b)] (optional) Let m ∈ N and consider a basis B = {1, (x − 5), (x − 5)2 , ..., (x − 5)m } of Pm (R).
What is the dual basis of B?
Proof. The dual basis of B is ϕ1 , ..., ϕm , where ϕj (p) =
(
i!, if i = j − 1
dj−1
i
.
dxj−1 (x − 5) x=5 =
0, if i 6= j − 1
dj−1
1
(j−1)! dxj−1
[p]x=5 for each j. This is because
[3F14] (optional) Define T : P(B) → P(B) by (T p)(x) = x2 p(x) + p00 (x) for x ∈ R.
(a) Suppose ϕ ∈ P(B)0 is defined by ϕ(p) = p0 (4). Describe the linear functional T 0 (ϕ) on P(B).
R1
(b) Suppose ϕ ∈ P(B)0 is defined by ϕ(p) = 0 p(x) dx. Evaluate (T 0 (ϕ)) (x3 ).
Proof.
(a) [T 0 (ϕ)] (p) = (ϕ◦T )(p) = ϕ x2 p(x) + p00 (x) = 2xp(x) + x2 p0 (x) + p000 (x) x=4 = 8p(4)+16p0 (4)+p000 (4).
R1
(b) [T 0 (ϕ)] (x3 ) = (ϕ ◦ T )(x3 ) = ϕ x5 + 6x = 0 x5 + 6x dx = 37
6 .
3