4
$\begingroup$

What can one conclude about a matrix, $M$, if its single eigenvalue is 1?

(I think the question is trying to demonstrate a contrast with the case where it is 0 instead of 1, in which we could conclude that the matrix is nilpotent.)

Can I conclude that the matrix is the identity matrix? Since $(M-I)^n=0$ by the Cayley-Hamilton theorem? Is there anything else?

Thanks.

  • 0
    The faq says that answering questions doesn't require any reputation at all.2011-11-08

2 Answers 2

2

Use the following (and the comments):

1) Any square matrix is similar to a triangular matrix.

2) Similar matrices have the same eigenvalues.

  • 0
    Yes, I missed your comment while typing this answer. We essentially said the same thing...2011-11-08
0

If you want compute fractional (or negative) powers of the matrix, you can't use diagonalization but must use the matrix-logarithm. (Try this for instance using the Pascal-matrix or the matrices of Stirling-numbers first and second kind)

  • 0
    Then you have many pleasant things to look forward to.2011-11-08