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We know that symmetric matrices are orthogonally diagonalizable and have real eigenvalues. Is the converse true? Does a matrix with real eigenvalues have to be symmetric?

A class of symmetric matrices, the positive definite matrices, have positive real eigenvalues. Is the converse true? Does a matrix with positive real eigenvalues have to be symmetric, positive-definite?

I think the answer to all this is "no", but I just wanted to confirm.

Thanks,

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    $\begingroup$ Consider [ 1 1; 0 1] $\endgroup$ – Nicolas Bourbaki Jul 2 '15 at 3:34
  • $\begingroup$ no, try to find example on upper trianguler matrix $\endgroup$ – Chiranjeev_Kumar Jul 2 '15 at 3:34
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Is the matrix

$$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

symmetric? It has only one positive eigenvalue of multiplicity two.

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    $\begingroup$ got it - thanks @Timbuc. $\endgroup$ – User001 Jul 2 '15 at 3:45
  • $\begingroup$ Hi @Timbuc, if we have a symmetric matrix, then it is positive-definite (or positive-semidefinite) if and only if its eigenvalues are positive (or non-negative). I guess this is somewhat useful? Perhaps computing eigenvalues to prove a symmetric matrix is PSD may be easier than showing that $x^tAx$ >0 for all x. What do you think? Thanks, $\endgroup$ – User001 Jul 2 '15 at 22:23
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    $\begingroup$ What is "somewhat useful"? To have positive definite matrices, or to know they are such by their eigenvalues? Both things are pretty important in certain things, dpeneding on your knowledge and7or to what you want to apply this stuff. $\endgroup$ – Timbuc Jul 2 '15 at 22:26
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    $\begingroup$ Either you calculate eigenvalues ... or ... you can also show all the main minors are positive (or the odd minors are negative in case of NSD matrices). $\endgroup$ – Timbuc Jul 2 '15 at 22:28
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    $\begingroup$ A quadratic form has to fulfill some basic requirements, and if the matrix isn't symmetric it doesn't. Try it with low dimensions, it's easy. $\endgroup$ – Timbuc Jul 2 '15 at 22:43
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No.

The simplest example is any upper triangular matrix whose diagonal entries are real:

$$\begin{pmatrix} 1 & 1-i & 2 \\ 0 & 2 & 3i \\ 0 & 0 & 3 \end{pmatrix}$$

has complex entries in the non-diagonal spots but its eigenvalues are $1,2,3$ and it is certainly not symmetric!

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  • $\begingroup$ ok got it - thanks, @xoque55. $\endgroup$ – User001 Jul 2 '15 at 3:46
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consider on this $$ \begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 3 \\ \end{bmatrix} $$

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  • $\begingroup$ You are welcome :) $\endgroup$ – Chiranjeev_Kumar Jul 2 '15 at 3:47
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The first question seems to be handled quite well by other people but I just want to add one more thing.

Positive-definiteness always starts from symmetry. So the converse can never be true.

That being said, it's only when we have the guarantee that the matrix is symmetric that we can conclude a matrix is positive-definite.

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  • $\begingroup$ Ok, awesome - thanks so much for adding this answer, @daeyounglim. So, positive eigenvalues does not imply positive-definiteness. $\endgroup$ – User001 Jul 2 '15 at 17:06

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