Exercise 2.3.14

Answers

1.

We can write

Bz = (\begin{matrix} \sum_{j = 1}^{p} a_{j} B_{1 j} \\ \sum_{j = 2}^{p} a_{j} B_{2 j} \\ ⋮ \\ \sum_{j = 1}^{p} a_{j} B_{nj} \end{matrix})

= \sum_{j = 1}^{p} a_{j} (\begin{matrix} B_{1 j} \\ B_{2 j} \\ ⋮ \\ B_{nj} \end{matrix}) = \sum_{j = 1}^{p} a_{j} v_{j} .

2.

This is the result of Theorem 2.13(b) and the previous exercise.

3.

This is instant result of the fact that

wA = A^{t} w^{t}

4.

This is also because

AB = B^{t} A^{t}

jephianlin

2011-06-27 00:00