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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!

Tim
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2 Answers2

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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.
Dominik
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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? – Tim Nov 25 '12 at 19:48
  • At least their eigenspaces must be orhogonal so that we can choose an orthonormal set of eigenvectors, to be more precise. – Dominik Nov 25 '12 at 19:50
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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)

Dominik
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  • Thanks! I mean triangular, not necessarily diagonal. – Tim Nov 25 '12 at 19:53
  • Is it possible to merge the answers? – c.p. Nov 25 '12 at 19:59
  • @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.) – Tim Nov 26 '12 at 12:56
  • Diagonalizabilitiy is a special case of triangularizability. – Dominik Nov 26 '12 at 13:03
  • @Dominik: I meant triangular, not triangularizability. – Tim Nov 26 '12 at 14:48
  • 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}}. – Dominik Nov 26 '12 at 17:36