Positive semidefinite but non diagonalizable real matrix - proof real parts of eigenvalues are non-negative I have a question about positive semidefinite matrices that are non diagonalizable. 
Example: 
\begin{equation}
A= \left(\begin{array}{cc}
2 &amp; 1\\
0 &amp; 2\\
\end{array}\right)
\end{equation}
Clearly the (real part of the) eigenvalues of $A$ are non-negative.

But how do I prove in general that the real part of the Eigenvalues of a positive semi-definite real matrix are non-negative?
(I have seen the proof where they use diagonalization of the matrix ($B=T^{-1}DT$) but this is not possible for all positive semi-definite real matrices.)
Of course, we assume that $A$ is a real matrix. The matrix $A$ may have non-real eigenvalues as this one $A=\begin{pmatrix}1&amp;1\\-1&amp;1\end{pmatrix}$. The correct result is:

If $A$ is real and for every $x\in\mathbb{R}^n$, $x^TAx\geq 0$, then the eigenvalues of $A$ have a non-negative real part.

cf. Does non-symmetric positive definite matrix have positive eigenvalues?

Now, I think that to call such a matrix $A$, a positive semi-definite matrix, is a very bad idea. Note that this condition is equivalent to  $A+A^T$ is $\geq 0$ in the usual sense. Moreover we encounter questions about this subject on MSE and often the OP does not even report that the studied matrix is not assumed to be symmetric!!
I guess PSD matrices are symmetric and a symmetric matrix is orthogonally diagonalizable.We allways consider symmetry because otherwise eigen values can be complex and then it loses the essence of psd.