# Are eigen spaces orthogonal?

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.

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.

• nice example. rank one matrix $\pmatrix{0 & 1\\0 & 1}$ will work too.
– abel
Jan 6, 2015 at 15:12
• @abel: No, that doesn't satisfy the condition of having $<N$ distinct eigenvalues. The matrix needs to be at least 3×3. Jan 6, 2015 at 15:51
• you are right. that seems a peculiar requirement, perhaps there is a reason for it.
– abel
Jan 6, 2015 at 15:55

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.

• A counterexample would be useful, in an answer. Jan 6, 2015 at 13:30
• am i correct about normal matrices ? Jan 6, 2015 at 13:30
• About normal matrices, yes, you are. Jan 6, 2015 at 13:31
• @sasha You are correct, normal matrices have orthogonal eigenspaces.
– 5xum
Jan 6, 2015 at 13:33
• @FedericoPoloni It has now been provided.
– 5xum
Jan 6, 2015 at 13:33