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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?


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You've seen what a Jordan block looks like? – J. M. Nov 8 '11 at 11:02
What are the eigenvalues of $\left[ {1\atop 1} {0\atop1}\right]$? – David Mitra Nov 8 '11 at 11:06
@J.M.: I just looked it up, so now I do! – impotent Nov 8 '11 at 11:07
@impotent Because $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]^2 = 0$, yet the matrix is not zero. – Sasha Nov 8 '11 at 11:14
@J.M. So here is my attempt at answering my own question... (I don't seem to be allowed to post this as an answer due to my lack of reputation...) Anyway, please correct me! $M$ can be similar to any upper triangular matrix with all diagonal entries equal to 1. because then the eigenvalues are all 1. Is this all I can say about $M$? Thanks again! – impotent Nov 8 '11 at 11:26

Use the following (and the comments):

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

2) Similar matrices have the same eigenvalues.

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Thanks, David. I tried answering the question --posted as a comment above, coz I am not allowed to post it as an answer... Am I right then? Thanks again! – impotent Nov 8 '11 at 11:28
Yes, I missed your comment while typing this answer. We essentially said the same thing... – David Mitra Nov 8 '11 at 11:30

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)

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Thanks, Gottfried! Unfortunately I have never come across the matrix logarithm, Pascal-matrix or the Stirling numbers... – impotent Nov 8 '11 at 11:53
Then you have many pleasant things to look forward to. – Gerry Myerson Nov 8 '11 at 12:02

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