Exercise 6.7.17

Answers

1.

We calculate

{(UT)}^{†}

and

U^{†}

T^{†}

separately. First we have

UT (x_{1}, x_{2}) = U (x_{1}, 0) = (x_{1}, 0) = T^{†} .

So we compute $T^{†}$ directly. We have $N (T)$ is the $y$ -axis. So $N {(T)}^{⊥}$ is the $x$ -axis. Since we have

T (1, 0) = (1, 0),

we know that

T^{†} (1, 0) = (1, 0)

and

T^{†} (0, 1) = (0, 0) .

Hence we have

{(UT)}^{†} (x_{1}, x_{2}) = T^{†} (x_{1}, x_{2}) = x_{1} T^{†} (1, 0) + x_{2} T^{†} (0, 1) = x_{1} T^{†} (1, 0) = (x_{1}, 0) .

On the other hand, we also have $N (U)$ is the line $span {(1, - 1)}$ . So $N {(U)}^{⊥}$ is the line $span {(1, 1)}$ . Since we have

U (1, 1) = (2, 0),

we know that

U^{†} (1, 0) = \frac{1}{2} (1, 1)

and

U^{†} (0, 1) = (0, 0) .

Hence we have

U^{†} (x_{1}, x_{2}) = x_{1} U^{†} (1, 0) + x_{2} U^{†} (0, 1) = (\frac{x_{1}}{2}, \frac{x_{1}}{2}) .

Finally we have

T^{†} U^{†} (x_{1}, x_{2}) = T^{†} (\frac{x_{1}}{2}, \frac{x_{1}}{2}) = (\frac{x_{1}}{2}, 0) \neq {(UT)}^{†} (x_{1}, x_{2}) .

2.

Let

A = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})

and

B = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

. By the previous argument, we have

A^{†} = (\begin{matrix} \frac{1}{2} & 0 \\ \frac{1}{2} & 0 \end{matrix})

and

B^{†} = B

. Also, we have

AB = B

and so

{(AB)}^{†} = B^{†} = B

jephianlin

2011-06-27 00:00