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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?

Thanks.

impotent
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    You've seen what a Jordan block looks like? – J. M. ain't a mathematician Nov 08 '11 at 11:02
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    What are the eigenvalues of $\left[ {1\atop 1} {0\atop1}\right]$? – David Mitra Nov 08 '11 at 11:06
  • @J.M.: I just looked it up, so now I do! – impotent Nov 08 '11 at 11:07
  • @DavidMitra: Ah, 1's only also. Hmm so it is not necessarily the identity matrix. But then what can I say about $M$? – impotent Nov 08 '11 at 11:08
  • May I encourage you to answer your own question (as soon as the software allows you to)? :) You did say that you now know what a Jordan block is; imagine then a matrix whose diagonal blocks are either identity matrices or Jordan blocks of various sizes... – J. M. ain't a mathematician Nov 08 '11 at 11:09
  • Incidentally, could someone explain why the Cayley-Hamilton argument didn't work? – impotent Nov 08 '11 at 11:10
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    @impotent Because $\left[ \begin{array}{cc} 0 & 1 \ 0 & 0 \end{array} \right]^2 = 0$, yet the matrix is not zero. – Sasha Nov 08 '11 at 11:14
  • @impotent: Yet another way of saying the same thing: When $\lambda=0$ was the sole eigenvalue of $A$, you were shown the conclusion that $A$ is nilpotent. If $\lambda=1$ is the only eigenvalue of $A$, then zero is the sole eigenvalue of $A-I$. Therefore ... – Jyrki Lahtonen Nov 08 '11 at 11:24
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    @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 08 '11 at 11:26
  • @JyrkiLahtonen: Interesting way to look at it! Thanks! – impotent Nov 08 '11 at 11:30
  • The faq says that answering questions doesn't require any reputation at all. – Gerry Myerson Nov 08 '11 at 12:03

2 Answers2

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Use the following (and the comments):

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

2) Similar matrices have the same eigenvalues.

David Mitra
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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 08 '11 at 11:28
  • Yes, I missed your comment while typing this answer. We essentially said the same thing... – David Mitra Nov 08 '11 at 11:30
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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)

Gottfried Helms
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