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Linear Algebra

Exercise 6.8.8

Answers

1.

Let

β = {v_{i}}

. We know that

{(ψ_{β} (H))}_{ij} = H (v_{i}, v_{j})

. So we have

{(ψ_{β} (c H_{1} + H_{2}))}_{ij} = (c H_{1} + H_{2}) (v_{i}, v_{j})

= c H_{1} (v_{i}, v_{j}) + H_{2} (v_{i}, v_{j}) = c {(ψ_{β} (H_{1}))}_{ij} + {(ψ_{β} (H_{2}))}_{ij} .

2.

The form

H^{'} (u, v) : = u^{t} Av

is a bilinear form when

u, v \in 𝔽^{n}

. We know that

ϕ_{β}

is an isomorphism from

V

to

𝔽^{n}

. This means that

H = \hat{ϕ_{β}^{- 1}} (H^{'})

is a bilinear form by Exercise 6.8.7.

3.

Let

β = {v_{i}}

. And let

x = \sum_{i = 1}^{n} a_{i} v_{i}, y = \sum_{i = 1}^{n} b_{i} v_{i} .

Thus we have

H (x, y) = H (\sum_{i = 1}^{n} a_{i} v_{i}, \sum_{i = 1}^{n} b_{i} v_{i})

\sum_{i, j} a_{i} b_{i} H (v_{i}, v_{j}) = {[ϕ_{β} (x)]}^{t} A [ϕ_{β} (y)] .

jephianlin

2011-06-27 00:00

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