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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.

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    You've seen what a Jordan block looks like?2011-11-08
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    What are the eigenvalues of $\left[ {1\atop 1} {0\atop1}\right]$?2011-11-08
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    @J.M.: I just looked it up, so now I do!2011-11-08
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    @DavidMitra: Ah, 1's only also. Hmm so it is not necessarily the identity matrix. But then what can I say about $M$?2011-11-08
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    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...2011-11-08
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    Incidentally, could someone explain why the Cayley-Hamilton argument didn't work?2011-11-08
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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.2011-11-08
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    @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 ...2011-11-08
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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!2011-11-08
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    @JyrkiLahtonen: Interesting way to look at it! Thanks!2011-11-08
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    The faq says that answering questions doesn't require any reputation at all.2011-11-08

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