Exercise 6.3.13

Answers

1.

x \in N (T^{∗} T)

we have

T^{∗} T (x) = 0

and

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

This means that $T (x) = 0$ and $x \in N (T)$ . Conversely, if $x \in N (T)$ , we have $T^{∗} T (x) = T^{∗} (0) = 0$ and so $x \in N (T^{∗} T)$ .

On the other hand, since the dimension is finite, we have

R (T^{∗} T) = N {(T^{∗} T)}^{⊥} = N {(T)}^{⊥} = R (T^{∗})

by the previous exercise. Hence we have

rank (T^{∗} T) = rank (T^{∗}) = rank (T)

by the next argument.

2.

For arbitrary matrix

A

, denote

\bar{A}

to be the matrix consisting of the conjugate of entris of

A

. Thus we have

A^{∗} = \bar{A^{t}}

. We want to claim that

rank (A) = rank (A^{∗})

first. Since we already have that

rank (A) = rank (A^{t})

, it’s sufficient to show that

rank (A) = rank (\bar{A})

. By Theorem 3.6 and its Corollaries, we may just prove that

{v_{i}}_{i \in I}

is independent if and only if

{\bar{v_{i}}}_{i \in I}

is independent, where

\bar{v_{i}}

means the vector obtained from

v_{i}

by taking conjugate to each coordinate. And it comes from the fact

\sum_{i \in I} a_{i} v_{i} = 0

if and only if

\sum_{i \in I} \bar{a_{i}} \bar{v_{i}} = \bar{\sum_{i \in I} a_{i} v_{i}} = 0 .

Finally, by Theorem 6.10 we already know that ${[T]}_{β}^{∗} = {[T^{∗}]}_{β}$ for some basis $β$ . This means that $rank (T) = rank (T^{∗})$ . And so

rank (T T^{∗}) = rank (T^{∗∗} T^{∗}) = rank (T^{∗}) = rank (T) .

3.

It comes from the fact

L_{A^{∗}} = {(L_{A})}^{∗}

jephianlin

2011-06-27 00:00