Exercise 6.4.22

Answers

1.

For brevity, denote

V_{1}

and

V_{2}

to be the spaces with inner products

⟨ ⋅,⋅ ⟩

and

{⟨ ⋅,⋅ ⟩}^{'}

respectly. Define

f_{y} (x) = {⟨ x, y ⟩}^{'}

be a function from

V_{1}

𝔽

. We have that

f_{y} (x)

is linear for

x

V_{1}

. By Theorem 6.8 we have

f_{y} (x) = ⟨ T (x), y ⟩

for some unique vector

T (x)

. To see

T

is linear, we may check that

⟨ T (x + z), y ⟩ = ⟨ x + z, y ⟩ = ⟨ x, y ⟩ + ⟨ z, y ⟩

= ⟨ T (x), y ⟩ + ⟨ T (z), y ⟩ = ⟨ T (x) + T (z), y ⟩

and

⟨ T (cx), y ⟩ = ⟨ cx, y ⟩ = c ⟨ x, y ⟩

c ⟨ T (x), y ⟩ = ⟨ cT (x), y ⟩

for all $x$ , $y$ , and $z$ .

2.

First, the operator

T

is self-adjoint since

⟨ x, T^{∗} (y) ⟩ = ⟨ T (x), y ⟩ = {⟨ x, y ⟩}^{'}

= \bar{⟨ y, x ⟩} = \bar{⟨ T (y), x ⟩} = ⟨ x, T (y) ⟩

for all $x$ and $y$ . Then $T$ is positive definite on $V_{1}$ since

⟨ T (x), x ⟩ = {⟨ x, x ⟩}^{'} > 0

if $x$ is not zero. Now we know that $0$ cannot be an eigenvalue of $T$ . So $T$ is invertible. Thus $T^{- 1}$ is the unique operator such that

⟨ x, y ⟩ = {⟨ T^{- 1} (x), y ⟩}^{'} .

By the same argument, we get that $T^{- 1}$ is positive definite on $V_{2}$ . So $T$ is also positive definite by Exercise 6.4.19(c).

jephianlin

2011-06-27 00:00