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Exercise 6.3.15

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1.

Let

y \in W

be given. We may define

g_{y} (x) = {⟨ T (x), y ⟩}_{2},

which is linear since $T$ is linear and the first component of inner product function is also linear. By Theorem 6.8 we may find an unique vector, called $T^{∗} (y)$ , such that

{⟨ x, T^{∗} (y) ⟩}_{1} = {⟨ T (x), y ⟩}_{2}

for all $x$ . This means $T^{∗} (y)$ is always well-defined. It’s unique since

{⟨ x, T^{∗} (y) ⟩}_{1} = {⟨ x, U (y) ⟩}_{1}

for all $x$ and $y$ implies that $T^{∗} = U$ .

Finally, it’s also linear since

\begin{array}{l} {⟨ x, T^{∗} (y + cz) ⟩}_{1} & = {⟨ T (x), y + cz ⟩}_{2} \\ = {⟨ T (x), y ⟩}_{2} + \bar{c} {⟨ T (x), z ⟩}_{2} \\ = {⟨ x, T^{∗} (y) ⟩}_{1} + \bar{c} {⟨ x, T^{∗} (z) ⟩}_{1} \\ = {⟨ x, T^{∗} (y) ⟩}_{1} + {⟨ x, c T^{∗} (z) ⟩}_{1} \\ = {⟨ x, T^{∗} (y) + c T^{∗} (z) ⟩}_{1} \end{array}

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

2.

Let

β = {v_{1}, v_{2}, \dots, v_{m}}

and

γ = {u_{1}, u_{2}, \dots, u_{n}} .

Further, assume that

T (v_{j}) = \sum_{i = 1}^{n} a_{ij} u_{i} .

This means that ${[T]}_{β}^{γ} = {a_{ij}}$ .

On the other hand, assume that

T^{∗} (u_{j}) = \sum_{i = 1}^{n} c_{ij} v_{i} .

And this means

\bar{c_{ij}} = {⟨ v_{i}, T^{∗} (u_{j}) ⟩}_{1} = {⟨ T (v_{i}), u_{j} ⟩}_{2} = a_{ji}

and ${[T^{∗}]}_{γ}^{β} = {({[T]}_{β}^{γ})}^{∗}$ .

3.

It comes from the same reason as Exercise 6.3.13(b).

4.

See

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

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

5.

T (x) = 0

we have

T^{∗} T (x) = T^{∗} (0) = 0

. If

T^{∗} T (x) = 0

we have

0 = ⟨ x, T^{∗} T (x) ⟩ = ⟨ T (x), T (x) ⟩

and hence $T (x) = 0$ .

jephianlin

2011-06-27 00:00

Exercise 6.3.15

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