Now I've got the right word: multiplicity of an eigenvalue, not order or rank.
How about part a)? How to proceed, or am I even on the right track?
I can only say that I don't know or understand. I could guess that because of what you said about the basises of an orthogonal matrix, then an...
I'm sorry. I probably meant order, not rank.
det(M-tI)=(-1)^3(t-\lambda _1)(t-\lambda _2)(t-\lambda _3)
and only one of the eigenvalues would be 1?
For the a) I've managed to get the following. I don't know if I'm supposed to apply the fact that I'm dealing with 3 x 3 matrix so that...
a) Let M be a 3 by 3 orthogonal matrix and let det(M)=1. Show that M has 1 as an eigenvalue. Hint: prove that det(M-I)=0.
I think I'm supposed to begin from the fact that
det(M)=1=det(I)=det(MTM) and from there reach det(M-I)=0 which of course would mean that there's an eigenvalue of 1 as...