Exercise 6.7.26

Answers

By Theorem 6.26, we know [T]β′γ′ = Σ for some orthonormal bases β′ and γ′, where Σ is the matrix in Theorem 6.27. In this case we know that

([T]β′γ′ )† = Σ† = [T†] γ′β′ .

Now for other orthonormal bases β and γ. We know that

[T]βγ = [I] γ′γΣ[I] ββ′

is a singular value decomposition of [T]βγ since both [I]γ′γ and [I]ββ′ are unitary by the fact that all of them are orthonormal. Hence we have

([T]βγ)† = [I] β′βΣ†[I] γγ′ = [I]β′β[T†] γ′β′ [I]γγ′ = [T†] γβ.
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2011-06-27 00:00
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