Assume you have a non-symmetric real square matrix all of whose eigenvalues are real. Can anything be said about it? Is it unitarily equivalent to a symmetric matrix?
EDIT: Is it at least similar to a symmetric matrix?
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3$\begingroup$ Re the last question: Clearly not in general. Consider a Jordan block of size greater than $1$ with one real eigenvalue. $\endgroup$– Geoff RobinsonCommented Feb 10, 2015 at 21:31
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3$\begingroup$ Can anything be said? Well, if you choose the $n^2$ matrix elements from independent normal distributions, the probability that all eigenvalues are real is $2^{-n(n-1)/4}$ (a result due to Ginibre). $\endgroup$– Carlo BeenakkerCommented Feb 10, 2015 at 21:37
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$\begingroup$ @ Geoff Robinson: Right, sorry. I have corrected my question. $\endgroup$– Delio MugnoloCommented Feb 10, 2015 at 21:49
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1$\begingroup$ @CarloBeenakker How can it be similar to a symmetric ($=$ diagonalizable) matrix if it has large Jordan blocks? Say, $[[1,1],[0,1]]$ is certainly not similar to $I$. $\endgroup$– Alex DegtyarevCommented Feb 10, 2015 at 22:44
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1$\begingroup$ @Carlo Beenakker: In what sense are you using the expression similaar? I thought $A$ and $B$ being similar meant $B = TAT^{-1}$ for some invertible matrix $T$. $\endgroup$– Geoff RobinsonCommented Feb 10, 2015 at 22:45
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2 Answers
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Well there is the following. Consider $$ \begin{equation} A = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right) \end{equation} $$ The matrix $A$ is non-symmetric with eigenvalues $\{0\}$. If $A$ were similar to a symmetric matrix $M$, then $A$ would be diagonalizable (because $M$ is), and we would have $A = 0$.
We can also look at $$ \begin{equation} B = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right) \end{equation} $$ The eigenvalues of $B$ are $\{1\}$. If $B$ were similar to a symmetric matrix, then $B$ would equal the identity.
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Call $A$ this matrix. If you assume in addition that $A$ is diagonalisable, then $A$ is a product of two symmetric matrices $S_+S$ where $S_+$ is positive definite and the signs of the eigenvalues of $S$ are the signs of the eigenvalues of $A$.
Of course, you don't expect that $A$ be unitarily (= orthogonaly here) similar to a symmetric matrix, because it would be itself symmetric.
As said by many, if $A$ is not diagonalisable, it cannot be similar to a symmetric matrix.
Finally, Theorem 4.1.7 in R. Horn & C. Johnson says that every real matrix is a product of two hermitian matrices, but I don't know whether you may take real factors.
answered Feb 10, 2015 at 22:53
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$\begingroup$ See H Stenzel: ``Über die Darstellbarkeit einer Matrix als Produkt von zwei symmetrischen Matrizen, als Produkt von zwei alternierenden Matrizen und als Produkt yon einer symmetrisehen und einer alternienden Matrix''. mathematische zeitschrift, volume 15, Number 1, 1922. If I understood it right (my german very old...), it can be written as the product of two symmetrical matrices, one of it can be taken positive definite. $\endgroup$ Commented Feb 11, 2015 at 17:48