Real part of a complex inner product is zero implies the imaginary part is also zero - why?

Question

This very useful discussion/proof of the spectral theorem states one thing that still confuses me - as an instrumental part of their proof, they suggest that for any self-adjoint linear operator $S$ on a finite vector space $V$, the following hold:

First defining: $$\varphi(v)=\langle S(v),v\rangle\\v_0=\operatorname{argmax}\varphi(v),v\in V:\langle v,v\rangle=1$$

I hope my notation was correct there - $\varphi$ attains a maximum on the unit sphere at $v_0$, is what that is trying to say.

They then prove the first line of the following, which I'm happy with, but assume the second line - which I'm not happy with!

$$\langle v_0,u\rangle=0\implies\Re\langle S(v_0),u\rangle=0\\\Re\langle S(v_0),u\rangle=0\implies\langle S(v_0),u\rangle=0$$

Is it true in general that if a complex inner product has zero real part, it is necessarily zero? Or is this simply a detail to do with the particular definitions of $S$ ($S$ as self-adjoint, or linear perhaps?), $v_0$ and $u$?

Answer 1

They are not claiming that $\operatorname{Re}\langle S(v_0),u\rangle$ implies $\langle S(v_0),u\rangle=0$. Instead, they actually explain how they are deducing that $\langle S(v_0),u\rangle=0$, namely by applying the first line "to all the vectors $e^{i\theta}u$". Or, more simply, since $e^{i\theta}u$ is just some $\mathbb{R}$-linear combination of $u$ and $iu$, we will just use it for $iu$. In other words, suppose that $$\langle v_0,u\rangle=0.$$ Then we also know that $$\langle v_0,iu\rangle=0.$$ But now we can apply the first line with $iu$ in place of $u$ to deduce that $$\operatorname{Re}\langle S(v_0),iu\rangle=0.$$ Now $\langle S(v_0),iu\rangle=-i\langle S(V_0),u\rangle$ so its real part is just the imaginary part of $\langle S(v_0),u\rangle$. So we know not only that the real part of $\langle S(v_0),u\rangle$ is $0$ but also that its imaginary part is $0$, so $\langle S(v_0),u\rangle=0$.

(This is a common trick: you prove a fact about all vectors, and then you apply that fact to all the different scalar multiples of one particular vector to reach a stronger conclusion about that one vector.)

Answer 2

(1.11) states $$\langle v_0, u \rangle = 0 \implies \Re \langle Sv_0, u \rangle = 0.$$ We will apply (1.11) multiple times to obtain the stronger result $$\langle v_0, w \rangle = 0 \implies \langle Sv_0, w \rangle = 0.$$

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Let $w$ be some vector and suppose $\langle v_0, w \rangle = 0$. Let $u = e^{i\theta} w$. Then $\langle v_0, u \rangle = \overline{e^{i\theta}} \langle v_0 , w \rangle = 0$. Applying (1.11) implies $$0 = \Re \langle Sv_0, u \rangle = \Re (\overline{e^{i\theta}} \langle Sv_0, w \rangle) = \cos(\theta) \Re\langle Sv_0, w \rangle + \sin(\theta) \Im\langle Sv_0, w \rangle.$$ This holds for any choice of $\theta$, so we must have $\Re \langle Sv_0, w \rangle = \Im \langle Sv_0, w \rangle = 0$. (Or as Eric Wofsey demonstrates, it suffices to check $\theta = 0$ and $\theta = \pi/2$.)