TITLE: Are eigen spaces orthogonal?
QUESTION [11 upvotes]: Let $A$ be a $N$ x $N$ matrix which has $k < N$ distinct eigenvalues. Are eigenspaces corresponding to different eigenvalues orthogonal in general ? I know it is true if $A$ is normal matrix. But can't prove in general.

REPLY [14 votes]: Counterexample: $\begin{pmatrix}1&0&0\\0&1&1\\0&0&2\end{pmatrix}$ has eigenspaces $\{(t,u,0)^T\}$ with eigenvalue $1$ and $\{(0,t,t)^T\}$ with eigenvalue $2$, and they are not orthogonal.

REPLY [8 votes]: You can't prove it in general because it's not true. In fact, for any linearly independent set of vectors $v_1,v_2,\dots, v_n\in\mathbb R^n$, you can define a matrix
$$P=[v_1,v_2,\dots,v_n]$$
and a matrix $D$ which is a diagonal matrix with pairwise distinct diagonal entries $\lambda_1, \lambda_2,\dots, \lambda_n$. 
Now, you know that $$(PDP^{-1})v_i = PD(P^{-1}v_i) = PDe_i = \lambda_i Pe_i = \lambda_i v_i.$$
This means that the vectors $v_1,\dots, v_n$ are eigenvectors, each spanning its distinct eigenspace (because the eigenvalues are pairwise distinct), and they are not, in general, orthogonal.