Matrices with real spectrum 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? 
 A: 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.
A: 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.
