orthogonal basis of eigenvector Let $A:\mathbb R^n \to \mathbb R^n $ be a diagonalizable matrix. From the definition, this means that there exist a basis of eigenvectors. Using the Gram.Schmidt Algorithm, one can prove that for every subspace of $\mathbb R^n$ there exist an orthogonal basis, but maybe this  basis is not an eigenvector basis. I want to know if for diagonalizable matrix, there always exist an orthogonal basis that are also eigenvectors of A. I think that's not true, but I can't find some example. I know at least that eigenvectors with different eigenvalues are orthogonal. 
Thanks!
 A: Let $A=\begin{bmatrix}1 & 1 \\ 0 & 2\end{bmatrix}$. $A$ is not normal, hence it cannot be unitarily diagonalized. However, it has distinct eigenvalues, hence it can be diagonalized.
To elaborate: Since the eigenvalues are distinct, the eigenspaces are one dimensional. It is easy to check that $\ker (A-I) = \operatorname{sp} \{ (1,0)^T \}$ and $\ker (A-2I) = \operatorname{sp} \{ (1,1)^T \}$, and these two spaces are not orthogonal. Hence an orthogonal basis of eigenvectors cannot exist.
A: Suppose that $\mathbb{R}^n$ has an orthogonal basis consisting of eigenvectors $v_1,v_2,\dots,v_n$ of the matrix $A$. We may as well assume the vectors have norm $1$. Thus, if we consider vectors as columns and the matrix
$$
U=\begin{bmatrix}v_1 & v_2 & \dots & v_n\end{bmatrix}
$$
and the diagonal matrix
$$
D=\begin{bmatrix}
d_1 & 0 & \dots & 0 \\
0 & d_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & d_n
\end{bmatrix}
$$
where $Av_k=d_kv_k$ $(k=1,2,\dots,n)$, we have
$$
A=UDU^{-1}.
$$
Since $U$ is orthogonal, we have that $U^{-1}=U^T$, the transpose of $U$. Now let's compute
$$
AA^T=(UDU^T)(UDU^T)^T=UDU^TUD^TU^T=UDD^TU^T
$$
and note that $DD^T=D^TD$, because they are diagonal (indeed $D=D^T$). So, with similar steps we come to the conclusion that
$$
AA^T=A^TA
$$
which is a necessary condition for $\mathbb{R}^n$ having an orthogonal basis consisting of eigenvectors of $A$.
It turns out that this condition, along with the fact that all the eigenvalues of $A$ are real, is also sufficient, but the proof is quite long. In particular symmetric matrices are diagonalizable (with an orthogonal diagonalizing matrix).

Over the complex numbers the condition $AA^H=A^HA$ ($A^H$ is the conjugate transpose) is necessary and sufficient for $\mathbb{C}^n$ having an orthogonal basis consisting of eigenvectors of $A$, because the existence of the eigenvalues over $\mathbb{C}$ is not a problem; the proof is the same with minor modifications.
