Exercise 6.11.3

Answers

1.
Check that A∗A = AA∗ = I. So A is orthogonal. Hence it’s a reflection by Theorem 6.45 since its determinant is − 1.
2.
Find the subspace {x : Ax = x}. That is, find the null space of A − I. Hence the axis is
span ⁡ {(3,1)}.
3.
Compute det ⁡ (B) = −1. Hence we have det ⁡ (AB) = det ⁡ (BA) = 1. By Theorem 6.45, both of them are rotations.
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2011-06-27 00:00
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