Solutions
Homework 4
19. Suppose T ∈ L(V ) and U is a subspace of V . Prove that U is invariant under T if
and only if PU T PU = T PU .
Proof. Assume that U is invariant under T , and let v ∈ V . Then since PU (v) ∈ U , it
follows that T PU (v) ∈ U . Since PU leaves vectors in U unchanged
PU T PU (v) = T PU (v).
Conversely, assume that PU T PU = T PU . Then for u ∈ U there are vectors u0 ∈ U
and u⊥ ∈ U ⊥ such that T (u) = u0 + u⊥ . Now
PU T PU (u) = T PU (u),
PU T (u) = T (u),
PU (u0 + u⊥ ) = u0 + u⊥ ,
u0 = u0 + u⊥ ,
and this implies that u⊥ = 0. Therefore T (u) = u0 ∈ U , and thus U is invariant
under T .
24. Find a polynomial q ∈ P2 (R) such that
Z 1
1
p
p(x)q(x)dx,
=
2
0
for every p ∈ P2 (R).
Solution: Let T : P2 (R) → R be given by T (p) = p 12 . Multiplication in R is an
inner product, therefore there is an adjoint map, T : R → P2 (R) defined by
T (p)r = hp, T ∗ (r)i,
for all real numbers, r, and 2nd degree polynomials, p. In particular,
Z 1
1
∗
p(x)T ∗ (1)(x)dx,
= T (p) = hp, T (1)i =
p
2
0
showing that q = T ∗ (1) is
√the desired polynomial.√Using the orthonormal basis from
6.10, e1 (x) = 1, e2 (x) = 3(2x − 1), and e3 (x) = 5(6x2 − 6x + 1),
T ∗ (1) = hT ∗ (1), e1 ie1 + hT ∗ (1), e2 ie2 + hT ∗ (1), e3 ie3 ,
= T (e1 )e1 + T (e2 )e2 + T (e3 )e3 ,
1
1
1
= e1
e1 + e2
e2 + e3
e3 ,
2
2
2
√
5
= e1 +
e3 .
2
Therefore q(x) = 30x2 − 30x + 6.
1
2
29. Suppose T ∈ L(V ) and U is a subspace of V . Prove that U is invariant under T if
and only if U ⊥ is invariant under T ∗ .
Proof. Assume that U is invariant under T , and let v ∈ U ⊥ . Then for all u ∈ U ,
hu, T ∗ (v)i = hT (u), vi = 0.
Therefore T ∗ (v) ∈ U ⊥ , and U ⊥ is invariant under T ∗ .
Conversely, assume that U ⊥ is invariant under T ∗ . By the above argument,
U = (U ⊥ )⊥ is invariant under T = (T ∗ )∗ .
30. Suppose T ∈ L(V, W ). Prove that
(a) T is injective if and only if T ∗ is surjective.
(b) T is surjective if and only if T ∗ is injective.
Proof. (a) Since (null T )⊥ = range T ∗ , null T = {0} if and only if range T ∗ = V .
Therefore T is injective if and only if T ∗ is surjective.
(b) Since (range T )⊥ = null T ∗ , range T = W if and only if null T ∗ = {0}. Therefore T is surjective if and only if T ∗ is injective.
31. Prove that
dim null T ∗ = dim null T + dim W − dim V,
and
dim range T ∗ = dim range T,
for every T ∈ L(V, W ).
Proof. Since (null T )⊥ = range T ∗ ,
dim V − dim null T = dim range T ∗ .
Substituting
dim V = dim null T + dim range T,
gives the equation,
dim range T = dim range T ∗ .
This and the equation
dim W = dim null T ∗ + dim range T ∗ ,
implies that
dim V − dim W = dim null T + dim range T − dim null T ∗ − dim range T ∗ ,
= dim null T − dim null T ∗ .
Therefore
dim null T ∗ = dim null T + dim W − dim V.
3
4.2.1. Verify
(a) (T1 + T2 )∗ = T1∗ + T2∗ .
(b) (aT )∗ = āT ∗ ; (ST )∗ = T ∗ S ∗ .
Solution: For all x and y,
hT1 (x) + T2 (x), yi = hT1 (x), yi + hT2 (x), yi,
= hx, T1∗ (y)i + hx, T2∗ (y)i,
= hx, T1∗ (y) + T2∗ (y)i.
haT (x), yi = ahT (x), yi = ahx, T ∗ (y)i = hx, āT ∗ (y)i.
hST (x), yi = hT (x), S ∗ (y)i = hx, T ∗ S ∗ (y)i.