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 & 1\\
0 & 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.)
 A: Of course, we assume that $A$ is a real matrix. The matrix $A$ may have non-real eigenvalues as this one $A=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$. 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!!
A: 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.
A: The following answer is copied and modified from achille hui's answer which can be found following the link in loup blanc's answer (Does non-symmetric positive definite matrix have positive eigenvalues?).
For the sake of completeness, I copied it here:
Let $A \in M_{n}(\mathbb{R})$ be any (non-symmetric) real $n\times n$ matrix but "positive semi-definite" in the sense that: 
$$\forall x \in \mathbb{R}^n, x \ne 0 \implies x^T A x \geq 0$$
The eigenvalues of $A$ need not be positive. For an example:
$$\begin{pmatrix}1&1\\-1&1\end{pmatrix}$$
has eigenvalue $1 \pm i$. However, the real part of any eigenvalue $\lambda$ of $A$ is always positive.
Let $\mathbb{C} \in \lambda = \mu + \nu i$ where $\mu, \nu \in \mathbb{R}$ be an eigenvalue of $A$. Let $z \in \mathbb{C}^n$ be a right eigenvector associated with $\lambda$. Decompose $z$ as $x + iy$ where $x, y \in \mathbb{R}^n$.
$$(A - \lambda) z = 0 \implies \left((A - \mu) - i\nu\right)(x + iy) = 0
\implies \begin{cases}(A-\mu) x + \nu y = 0\\(A - \mu) y - \nu x = 0\end{cases}$$
This implies
$$x^T(A-\mu)x + y^T(A-\mu)y = \nu (y^T x - x^T y) = 0$$
and hence
$$\mu = \frac{x^TA x + y^TAy}{x^Tx + y^Ty} \geq 0$$
In particular, this means that all real parts of all eigenvalues $\lambda$ of $A$ are non-negative.
