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Linear Algebra (2015 5 ed) by Lay, p. 397. Theorem 7.1 involves only real numbers.

Let:

  1. $A^* = \bar{A}^T $.

  2. $v_i$ and $v_j$ be two eigenvectors of an Hermitian matrix H. Suppose that their respective eigenvalues i and j are different, i.e. $\lambda_i \neq \lambda_j$. This means $Hv_i = \lambda_iv_i$ and $Hv_j > = \lambda_jv_j \quad (3)$.

Take the Hermitian conjugate of $Hv_i = \lambda_iv_i$:

$\begin{align} (Hv_i)^* & = (\lambda_iv_i)^* \\ \implies v_i^* H^* & = > v_i^* \lambda_i^* \\ v_i^* H & = \qquad \qquad \text{ > because H* = H } \\ & = v_i^* \lambda_i \qquad \text{ because > http://math.stackexchange.com/q/462982/53259 } \end{align}$

Right-multiply the previous equation by $\color{green}{v_j}: \qquad > v_i^*H \color{green}{v_j} = v_i^* \lambda_i \color{green}{v_j} \qquad > (4)$

Left-multiply (3) by $\color{orangered}{v_i^*} : \quad > \color{orangered}{v_i^*}Hv_j = \color{orangered}{v_i^*} \lambda_jv_j > \qquad (5)$.

Equate the RHS of (4) and (5): $\quad v_i^* \lambda_i > \color{green}{v_j} = \color{orangered}{v_i^*} \lambda_jv_j \qquad > \blacksquare$.

I understand, and ask not about, the algebra. What's the proof strategy? For example, how can you divine when to take the Hermitian conjugate, what to multiply, and when to left- or right-multiply?

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  • $\begingroup$ Do not use "brook". Do you want to use "understand"? Say "understand". Do you want to say "put up with"? It appears unnecessarily rude, but if that's what you mean, say "put up with". $\endgroup$ – Andrés E. Caicedo Apr 21 '14 at 17:11
  • $\begingroup$ The whole point is that two (column) vectors $v,w$ in $\mathbb C^n$ are orthogonal iff $v^*w=0$. If $v,w$ are eigenvectors of a matrix $A$, then $Aw$ is a multiple of $w$, and $Av$ is a multiple of $v$. Now, unless $Av=0$ or $Aw=0$, this means that $v^*w=0$ iff $(Av)^*w=0$ iff $v^*(Aw)=0$ iff $(Av)^*(Aw)=0$. Since $A$ is Hermitian, that is, $A^*=A$, and since $(Av)^*=v^*A^*=v^*A$, all this suggests trying to compute these expressions ($(Av)^*w$, $v^*(Aw)$, $(Av)^*(Aw)$), and seeing whether something useful comes out of at least one of them. Once we have this idea, the rest is mechanics. $\endgroup$ – Andrés E. Caicedo Apr 22 '14 at 2:57
  • $\begingroup$ @AndresCaicedo: Thanks. I meant "put up with" but it does appear rude. Would you please explain why you advised against 'brook'? $\endgroup$ – Greek - Area 51 Proposal Apr 24 '14 at 9:25
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Let's write the problem in a better way. You have a Hermitian matrix $A$, two distinct eigenvalues $\lambda$ and $\mu$ and two eigenvectors $v\ne0$, $w\ne0$ such that $$ Av=\lambda v,\qquad Aw=\mu w. $$ In this first proof, the strategy is to work backwards. You want to prove that $v^*w=0$. So start with the assumption that $\lambda-\mu\ne0$; so $v^*w=0$ if and only if $(\lambda-\mu)v^*w=0$, that is $$ (\lambda v^*)w=\mu v^*w $$ On the LHS, but $Av=\lambda v$, so $\lambda v^*=v^*A^*$ (because $\lambda$ is real).
On the RHS, $\mu w=Aw$, so you're bound to prove that $$ (v^*A^*)w=v^*Aw $$ which immediately follows from $A^*=A$.

A different proof, but using the same techniques in a different order: start from $v^*Aw$ that you can rewrite as $$ v^*(Aw)=v^*(\mu w)=\mu (v^*w) $$ but also as $$ v^*A^*w=(Av)^*w=(\lambda v)^*w=\lambda(v^*w) $$ thus $$ \lambda(v^*w)=\mu (v^*w) $$ which can happen only if $\lambda=\mu$ or $v^*w=0$.

Remark on notation

With $v^*w$ I mean the product of the conjugate transpose (H-transpose, briefly) of $v$ by $w$, which is a $1\times 1$ matrix, which is the same as a number. The inner product we use here is exactly $v^*w$.

It's a question of conventions; $v^*w=0 \iff w^*v=0 \iff \color{orangered}{ ( } w^*v \color{orangered}{ )^T} =0 \iff v^T \bar{w}=0 $,
so orthogonality is the same with all three definitions.

Remark on the first proof

The assumption that $\lambda\ne\mu$ is clearly necessary, because there's no way to prove that two eigenvectors relative to the same eigenvalue are orthogonal.

Thus, if we want to prove $v^*w=0$, we can as well prove that $(\lambda-\mu)v^*w=0$, which allows to distribute over the subtraction.

Remark on the second proof

Here we use the other hypothesis, that is $A=A^*$. But this is not a necessary assumption, because the theorem is true as soon as $A$ is a normal matrix, that is $AA^*=A^*A$. We want a “skew-symmetric” expression and $v^*Aw=v^*A^*w$ is surely a first shot.

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  • $\begingroup$ Thank you. About the 2nd paragraph of your first proof, what does $v*w = 0$ mean? Does * denote conjugate transpose or dot product (v, w are orthogonal means $v \cdot w = 0$.) Also, how would you determine/divine/previse to start with the assumption $\lambda \neq \mu$? $\endgroup$ – Greek - Area 51 Proposal Apr 23 '14 at 9:56
  • $\begingroup$ About your second proof, how would you determine/divine/previse to write both $v*(Aw)$ and $v*A*w$ ? They feel "sibylline" to me. $\endgroup$ – Greek - Area 51 Proposal Apr 23 '14 at 9:57
  • $\begingroup$ +1. Thank you again. Regarding the notation, don't we need to prove only $v \cdot w = 0$ (ie v, w are orthogonal)? Please advise if I misconstrued you, but you are saying that we need to prove $\bar{v}^T \cdot w = 0$ instead? $\endgroup$ – Greek - Area 51 Proposal Apr 24 '14 at 9:24
  • $\begingroup$ @LePressentiment $v\cdot w=v^*w$ by definition (or it is $w^*v$, depending on conventions). $\endgroup$ – egreg Apr 24 '14 at 9:26
  • $\begingroup$ @LePressentiment Defining $v\cdot w=v^Tw$ does not define an inner product on a vector space over the complex numbers: you wouldn't have $v\cdot v$ real and nonnegative, for instance. I think you're confusing between real and complex vector spaces. $\endgroup$ – egreg Apr 24 '14 at 9:33

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