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As in the title. Also, if anyone knows if all Hermitian-symmetric matrices with distinct diagonal elements are diagonalizable, that'd be great to know. Thanks.

Edit: Never mind about the Hermitian one, answered here: Are singular matrices diagonalizable?

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  • $\begingroup$ According to the page on normal matrices and the spectral theorem, normal matrices are diagonalizable. Is a Toeplitz matrix a normal matrix? $\endgroup$ – bright-star Feb 27 '13 at 8:24
  • $\begingroup$ From the theory of Jordan decomposition, we know that if the geometric multiplicity of the eigenvalues of the matrix is less than the algebraic multiplicity, the matrix is not strictly diagonalizable. Perhaps there's something about Toeplitz matrices that can tell us about that? $\endgroup$ – bright-star Feb 27 '13 at 21:14
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    $\begingroup$ Hint: What is the first non-diagonalizable matrix you can think of? $\endgroup$ – David E Speyer Feb 28 '13 at 15:36
  • $\begingroup$ Oh! Putting the answer down later when I'm not in a hurry. Thanks :) $\endgroup$ – bright-star Feb 28 '13 at 18:01
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The answer is no. Consider a Toeplitz matrix of this form:

$A = \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{bmatrix}$

$A$ is a non-diagonalizable matrix with eigenvalues equal to $\lambda$, but it is not diagonalizable because the algebraic multiplicity (number of repeated eigenvectors) of $\lambda$ is 4, which is greater than its geometric multiplicity (number of linearly independent eigenvectors of $\lambda$), which is 1.

In fact, this is as diagonalized as this matrix is going to get, because this is the Jordan form of some defective matrix.

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    $\begingroup$ On the other hand, the special case of circulant matrices are always diagonalizable (simultaneously in fact, for a given order). $\endgroup$ – Erick Wong Mar 1 '13 at 1:13

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