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Every eigenvalue of a unitary matrix has absolute value 1. I was wondering whether a matrix whose eigenvalues all have absolute value 1 must be unitary?

Thanks!

2 Answers 2

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  1. No, the eigenvectors of a unitary matrix must also be orthogonal. So for example the matrix with Eigenvectors (1,0) and (1,1) with eigenvalues 1 and -1, respectively, is not unitary.
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    Thanks, but I think the eigenvectors of a unitary matrix are not necessarily orthogonal. If you are right, how would you characterize a unitary matrix via its eigenvalues and/or eigenvectors?2012-11-25
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    At least their eigenspaces must be orhogonal so that we can choose an orthonormal set of eigenvectors, to be more precise.2012-11-25
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2: Yes, if the algebraic multiplicity of all eigenvectors equal their geometric multiplicity, then the matrix is diagonalisable because the dimensions of the eigenspaces add up to $n$ so that you can choose $n$ linear independent eigenvectors (at least over an algebraically closed field)

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    Thanks! I mean triangular, not necessarily diagonal.2012-11-25
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    Is it possible to merge the answers?2012-11-25
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    @Dominik: I think there is some misunderstanding, is there? My question is about when matrices are triangular, and your reply is about when matrices are diagonalizable. I can't see the connection. (BTW, I am going to accept your other reply, and create a new post for the second question. You are welcome to move your reply there.)2012-11-26
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    Diagonalizabilitiy is a special case of triangularizability.2012-11-26
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    @Dominik: I meant triangular, not triangularizability.2012-11-26
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    Then the answer must be no, of course. For example you can consider {{1,2},{3,4}}^-1*{{1,0},{0,-1}}*{{1,2},{3,4}}={{-5,8}{3,5}}.2012-11-26