Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)

Solution 1:

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?

Solution 2:

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 $A$.

Solution 3:

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 : $$\sigma(u,v)=\sum_{k=0}^3 i^k\sigma(u+i^k v, u+i^kv). $$

Solution 4:

linear_algebra_done_right This is from linear algebra done right. btw, it's the greatest book about linear algebra I've ever seen!

Proof: Suppose $\alpha,\beta$ are distinct eigenvalues of $T$, with corresponding eigenvectors $u,v$. Thus, $Tu = \alpha u$ and $Tv = \beta v$. From 7.21 we have $T^*v = \bar \beta v.$ Thus, \begin{align} (\alpha - \beta)\langle u,v \rangle &= \langle \alpha u,v \rangle - \langle u, \bar \beta v \rangle \\ &= \langle Tu,v \rangle - \langle u, T^*v \rangle \\ & = 0. \end{align} Because $\alpha \neq \beta$, the equation above implies $\langle u,v \rangle = 0$. Thus, $u$ and $v$ are orthogonal, as desired.