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The problem statement is as follows:

Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.

I can certainly prove that this is the case, using the spectral theorem. The gist of my proof is presented below.

If possible, I would like to find a simpler proof. I was hoping that there might be some sort of manipulation along these lines, noting that $$ \langle Av_1,A v_2\rangle = \langle v_1,A^*Av_2\rangle = \langle v_1,AA^*v_2\rangle = \langle A^* v_1,A^* v_2 \rangle $$

Any ideas here would be appreciated.


My proof:

Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that $$ A v_{\lambda,i} = \lambda v_{\lambda,i} $$ Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$. We may write $ v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda} .$ We then have $$ 0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda} $$ So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$. Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$. From there, we have $$ \langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda} $$ the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.

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  • $\begingroup$ "Let $v_{\lambda,i}$ be an orthonormal basis of eigenvectors..." I'd guess that this might be something very close to what would need to be proved. $\endgroup$ – Algebraic Pavel May 3 '14 at 2:13
  • $\begingroup$ @PavelJiranek I was worried that I had used circular logic at some point. However, the existence of such a basis (i.e. the spectral theorem) comes directly from the Schur triangularization theorem, which says nothing about normal matrices in particular. $\endgroup$ – Omnomnomnom May 3 '14 at 19:11
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Assume $\;\lambda\neq \mu\;$ and

$$\begin{cases}Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v\\{}\\Aw=\mu w\implies A^*w=\overline\mu w\end{cases}$$

From this we get:

$$\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\{}\\ \langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}$$

and since $\;\overline\mu\neq\overline\lambda\;$ , we get $\;\langle v,w\rangle =0\;$

Question: Where did we use normality in the above?

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    $\begingroup$ How de we know that $v$ is also an eigenvector for $A^*$? $\endgroup$ – Berci May 2 '14 at 22:55
  • $\begingroup$ @Berci, read and think of the question at the end of the answer. $\endgroup$ – DonAntonio May 2 '14 at 22:56
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    $\begingroup$ From the other two answers this follows, of course, using that $A$ is normal. But can you show it directly? $\endgroup$ – Berci May 2 '14 at 22:57
  • $\begingroup$ Why directly? Use the definition and work it out. I usually don't give full answers and, after all, the OP is trying to get a more or less simpler proof to the claim than the one using the spectral theorem... $\endgroup$ – DonAntonio May 2 '14 at 23:00
  • $\begingroup$ I like it! Showing that $Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v$ seems to be the step in the spirit of what I was looking for. Thanks. $\endgroup$ – Omnomnomnom May 2 '14 at 23:14
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Specializing your identity to $v_1=v_2=v$, we get $\|Av\|=\|A^*v\|$. Hence $\ker A=\ker A^*$. Recalling that $\ker A^* = (\operatorname{ran} A)^\perp$ for general $A$, we conclude that the kernel and range of a normal matrix are mutually orthogonal.

It remains to apply the above conclusion to $A-\lambda I$ where $\lambda$ is an eigenvalue of $A$.

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  • $\begingroup$ Neat! Although DonAntonio's proof is more along the lines I was thinking of, this is a nice perspective. $\endgroup$ – Omnomnomnom May 2 '14 at 23:16
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I try to give another simple proof to $$T^*v=\bar{\lambda}v ~\text{ if }~ Tv=\lambda v$$ where $T$ is a normal operator on a Hilbert space $H$.

Suppose $V=\ker(T-\lambda I)$. Since $T^*$ communicate with $T$, $$T^*V\subset V.$$ Because $$\langle v,T^*v\rangle =\langle Tv,v\rangle =\langle \lambda v,v\rangle=\langle v,\bar{\lambda}v\rangle ~~\forall v \in V, $$ $\langle u,T^*v\rangle =\langle u, \bar{\lambda}v\rangle ~\forall u,v\in V$ by polarisation identity, and thus $T^*v=\bar{\lambda}v.$

REMARK: Let $\sigma:V\times V\to W$ be a sesquilinear form, where $V$ and $W$ are linear vector spaces over $\mathbb{C}$. The follwing formula is called Polarisation Identity : $$\sigma(u,v)=\sum_{k=0}^3 i^k\sigma(u+i^k v, u+i^kv). $$

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  • $\begingroup$ I don't really see how this answer addresses the question being asked. Also, the question being asked was answered 3 years ago. $\endgroup$ – Omnomnomnom Jun 12 '17 at 15:01
  • $\begingroup$ @Omnomnomnom This is why $Av=\lambda v\Rightarrow A^*v=\bar{\lambda}v$ using no spectral theorem which is the point to the question I think. $\endgroup$ – C.Ding Jun 13 '17 at 2:37
  • $\begingroup$ That is not the point to the question, actually. The point is to show that eigenvectors corresponding to different eigenvalues are necessarily orthogonal. $\endgroup$ – Omnomnomnom Jun 13 '17 at 2:58
  • $\begingroup$ Oh, sorry to disturb you. As I give some additional remarks to the right answer you have accepted. $\endgroup$ – C.Ding Jun 13 '17 at 3:04
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A normal matrix is unitarily similar to diagonal matrix.

$$A = UDU^{-1}$$ where $U$ is Unitary matrix.

Eigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. And matrix $D$ is Diagonal matrix with eigenvalues on diagonal.

Property: Columns of Unitary matrix are orthogonal.

So, columns of $U$ (which are eigenvectors of $A$) are orthogonal.

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