(adsbygoogle = window.adsbygoogle || []).push({}); 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(M^{T}M) and from there reach det(M-I)=0 which of course would mean that there's an eigenvalue of 1 as det(M-tI)=0 for any eigenvalue t.

I mean:

det(M)=1=det(I)=det(M^{T}M) => hard work => det(M-1I)=0 => t=1

b) Let M be a 3 by 3 orthogonal matrix and let det(M)=1. Show that either M=I or the eigenvalue 1 is of rank 1. Hint: M is diagonalizable.

I guess I know how to show the case where M=I:

det(M)=1=det(I)=det(P^{T}P)=det(P^{T}IP)=det(P^{T}MP) => M=I.

But that doesn't cover the fact that the rank of the eigenvalue could be 1.

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# Homework Help: Orthogonal matrix and eigenvalues

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