- #1
wormbox
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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 det(M-tI)=0 for any eigenvalue t.
I mean:
det(M)=1=det(I)=det(MTM) => 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(PTP)=det(PTIP)=det(PTMP) => M=I.
But that doesn't cover the fact that the rank of the eigenvalue could be 1.
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 det(M-tI)=0 for any eigenvalue t.
I mean:
det(M)=1=det(I)=det(MTM) => 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(PTP)=det(PTIP)=det(PTMP) => M=I.
But that doesn't cover the fact that the rank of the eigenvalue could be 1.