Question

Prove that, if T is an orthogonal transformation on R2 such that D(T) = -1, there exists an orthonormal basis for R2 such that the matrix of T
with respect to this basis is
(-1 0)
(0 1 ).

Answer 1

Thanks for your help in advance!!

Answer 2

Is D(T) the determinant?

Answer 3

Yessir

Answer 4

So far, all I have is since D(T) = -1 which does not equal 0, we have two linearly independent row vectors for T. (Dim T = 2.)

Answer 5

http://galletue.ing-mat.udec.cl/~camilo/material_udec/icm/Libros/Curtis,%20C.%20W.%20-%20Linear%20Algebra,%20An%20Introductory%20Approach.pdf

This is the online textbook, we're on page 270 and theorem 30.5 seems relevant.

Answer 6

@Traxter

Answer 7

What?

Answer 8

Would you have any idea on how to start this problem? Would we need to find the minimal polynomial?

Answer 9

if A =[ u v ] is a 2x2 orthogonal matrix then ||u|| =||v|| =1 and u.v = 0.

||u|| =1 implies u =[ cos(alpha), sin(alpha)]' for some alpha.

u.v =0 and ||v||=1 implies v = +- [-sin(alpha, cos(alpha)]´.

If v = [-sin(alpha), cos(alpha)]' then det(A) = 1, so this is not the case.

if v = [sin(alpha),-cos(alpha)]' then det(A) = -1 and therefore A = [ cos(alpha) sin(alpha); sin(alpha) -cos(alpha)] (matlab notation)

A is symmetric with eigenvalues lambda=-1,1 with corresponding eigenvectors p, q which are orthogonal and can be chosen to have length =1. The basis {p,q} is othonormal and the matrix that represents A with respect to this basis is [-1 0; 0 1].