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.


2 Answers 2


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.

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


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.

  • 3
    $\begingroup$ A counterexample would be useful, in an answer. $\endgroup$ Jan 6, 2015 at 13:30
  • $\begingroup$ am i correct about normal matrices ? $\endgroup$ Jan 6, 2015 at 13:30
  • 1
    $\begingroup$ About normal matrices, yes, you are. $\endgroup$
    – fonini
    Jan 6, 2015 at 13:31
  • $\begingroup$ @sasha You are correct, normal matrices have orthogonal eigenspaces. $\endgroup$
    – 5xum
    Jan 6, 2015 at 13:33
  • $\begingroup$ @FedericoPoloni It has now been provided. $\endgroup$
    – 5xum
    Jan 6, 2015 at 13:33

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