# Orthogonal matrix and eigenvalues

1. Aug 21, 2008

### wormbox

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.

2. Aug 21, 2008

### HallsofIvy

Staff Emeritus
What is the definition of the "rank" of an eigenvalue?

3. Aug 21, 2008

### wormbox

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 det(-M)=(-1)³det(M)...

$$\begin{array}{l} det(M)=1=det(I)\ \ \ |\cdot det(M^T-I) \Rightarrow \\ det(M)det(M^T-I)=det(I)det(M^T-I) \\ det(I-M)=det(M^T-I) \\ -det(M-I)=det(M^T-I) \\ det(M-I)=det(I-M^T) \end{array}$$

Last edited: Aug 21, 2008
4. Aug 21, 2008

### Avodyne

Since M is diagonalizable, let's work in a basis where it is diagonal. M is orthogonal, which means the transpose of M times M equals I (in any basis). What does this tell us about each eigenvalue?

5. Aug 21, 2008

### wormbox

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 orthogonal matrix has only linearly independent eigenvectors, which in turn would mean that all the eigenvalues are distinct.

6. Aug 21, 2008

### Dick

The eigenvalues are roots of a real polynomial. That means the eigenvalues are either real or come in complex conjugate pairs. Does that help?