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Exercise 6.5.30

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We may write

β = {v_{1}, v_{2}, \dots, v_{n}}

and

γ = {u_{1}, u_{2}, \dots, u_{n}} .

We have that $Q = {[I]}_{γ}^{β}$ . Now if $β$ is orthonormal, we may compute

= u_{j} = \sum_{i = 1}^{n} Q_{ij} v_{i} .

Thus we know that the inner product of $u_{s}$ and $u_{t}$ would be

⟨ u_{s}, u_{t} ⟩ = ⟨ \sum_{i = 1}^{n} Q_{is} v_{i}, \sum_{i = 1}^{n} Q_{it} v_{i} ⟩

= \sum_{i = 1}^{n} Q_{is} \bar{Q_{it}},

the value of inner product of the $s$ -th and the $t$ -th columns of $Q$ . So it would be $1$ if $s = t$ and it would be $0$ if $s \neq t$ . Finally the converse is also true since $Q^{∗} = {[I]}_{β}^{γ}$ is also an unitary matrix.

jephianlin

2011-06-27 00:00

Exercise 6.5.30

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