11
$\begingroup$

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.

$\endgroup$

2 Answers 2

14
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
8
$\begingroup$

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.

$\endgroup$
5
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.