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I know that symmetric matrices have real eigenvalues, and that non-symmetric matrices that are similar to symmetric matrices must also have real eigenvalues, but is the converse true?

That is, if a matrix has real eigenvalues, must there exist a similar matrix that is symmetric?

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No. Counterexample: $$ \pmatrix{0&1\\0&0} $$ is not similar to any symmetric matrix.

On the other hand, every diagonalizable matrix with real eigenvalues is similar to a symmetric matrix.

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  • $\begingroup$ Thanks for the counter-example. Indeed, I am indeed looking at diagonalizable matrices. Can you explain in more detail or sketch out a proof if it is short, or point me to a text that shows this? $\endgroup$
    – science404
    Aug 14, 2015 at 21:38
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    $\begingroup$ @science404 If a matrix is diagonalizable, it is similar to a diagonal matrix, which is symmetric $\endgroup$
    – angryavian
    Aug 14, 2015 at 21:38
  • $\begingroup$ Of course, thanks! $\endgroup$
    – science404
    Aug 14, 2015 at 21:39

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