Exercise 5.1.7

Answers

1.
We have that
[T]β = [I]γβ[T] γ[I]βγ

and

([I]γβ)−1 = [I] βγ.

So we know that

det ⁡ ([T]β) = det ⁡ ([I]γβ)det ⁡ ([T] γ)det ⁡ ([I]βγ) = det ⁡ ([T] γ).
2.
It’s the instant result from Theorem 2.18 and the Corollary after Theorem 4.7.
3.
Pick any ordered basis β and we have
det ⁡ (T−1) = det ⁡ ([T−1] β) = det ⁡ (([T]β)−1) = det ⁡ ([T] β)−1 = det ⁡ (T)−1.

The second and the third equality come from Theorem 2.18 and the Corollary after Theorem 4.7.

4.
By definition,
det ⁡ (TU) = det ⁡ ([TU]β) = det ⁡ ([T]β[U]β)
= det ⁡ ([T]β)det ⁡ ([U]β) = det ⁡ (T)det ⁡ (U).
5.
We have
det ⁡ (T − λIV ) = det ⁡ ([T − λI]β) = det ⁡ ([T]β − λ[I]β) = det ⁡ ([T]β − λI).
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2011-06-27 00:00
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