Exercise 6.1.23

Answers

1.

We have the fact that with the standard inner product

⟨ ⋅,⋅ ⟩

we have

⟨ x, y ⟩ = y^{∗} x

. So we have

⟨ x, Ay ⟩ = {(Ay)}^{∗} x = y^{∗} A^{∗} x = ⟨ A^{∗} x, y ⟩ .

2.

First we have that

⟨ A^{∗} x, y ⟩ = ⟨ x, Ay ⟩ = ⟨ Bx, y ⟩

for all $x, y \in V$ . By Theorem 6.1(e) we have $A^{∗} x = Bx$ for all $x$ . But this means that these two matrix is the same.

3.

Let

β = {v_{1}, v_{2}, \dots, v_{n}}

. So the column vectors of

Q

are those

v_{i}

’s. Finally observe that

{(Q^{∗} Q)}_{ij}

v_{i}^{∗} v_{j} = ⟨ v_{i}, v_{j} ⟩ = {\begin{matrix} 1 & if & i = j; \\ 0 & if & i \neq j . \end{matrix}

So we have $Q^{∗} Q = I$ and $Q^{∗} = Q^{- 1}$ .

4.

Let

α

be the standard basis for

𝔽^{n}

. Thus we have

{[T]}_{α} = A

and

{[U]}_{α} = A^{∗}

. Also we have that actually

{[I]}_{α}^{β}

is the matrix

Q

defined in the previous exercise. So we know that

{[U]}_{β} = {[I]}_{α}^{β} {[U]}_{α} {[I]}_{β}^{α} = QA Q^{- 1} = QA Q^{∗}

= {(Q A^{∗} Q^{∗})}^{∗} = {[I]}_{α}^{β} {[T]}_{α} {[I]}_{β}^{α} = {[T]}_{β} .

jephianlin

2011-06-27 00:00