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I am trying to prove a theorem related to PSD matrices, and have to make use of the Cauchy-Schwarz inequality with inner product not requiring definiteness.

The proof of Cauchy-Schwarz I have read either uses the definitess property of inner product, or it does not use it but only proves it in the real case.

I have found this, but I am having trouble extending the proof to the complex case. How do I deal with the Re{}?

Also, it didn't prove the condition for equality, and I would like to fill that in in my note.

Any guidance is appreciated.

Tim
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  • use am gm inequatlity – Aryan Kr. Sep 28 '24 at 13:42
  • Cauchy-Schwarz inequality holds in any vector space equipped with a semi-inner product $\langle\cdot,\cdot\rangle$. It requires semi-definiteness, not definiteness. For any two vectors $x$ and $y$, let $\langle x,y\rangle=re^{i\theta}$ in polar form and let $\omega=e^{-i\theta}$. Then $\langle x,\omega y\rangle=r=|\langle x,y\rangle|$. So, for any real number $t$, we have $$ 0\le\langle tx+\omega y,tx+\omega y\rangle =t^2\langle x,x\rangle+2t|\langle x,y\rangle|+\langle y,y\rangle. $$ Hence the discriminant $4|\langle x,y\rangle|^2-4\langle x,x\rangle\langle y,y\rangle$ is $\le0$. – user1551 Sep 28 '24 at 15:21
  • @user1551 The main trick (multiplying $y$ by $\omega$) was the same in my answer, where the second trick (sign of the discriminant) was avoided, and the "exceptional" case $\langle x,x\rangle=0$ was also solved. – Anne Bauval Sep 28 '24 at 15:39
  • you can sidestep polarization by letting $G$ be a $2\times 2$ Gram Matrix with $g_{i,j}=\langle x_i, x_j\rangle$; using a quadratic form argue it is Hermitian PSD so it is congruent to a very simple diagonal matrix which implies $\det\big(G\big)\geq 0$ which gives the result. [Or argue it can't have negative eigenvalues and all eigenvalues are real to get the determinant inequality] – user8675309 Sep 28 '24 at 16:26

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Let $V$ be a vector space and $\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{C}$ be a hermitian form (conjugate-linear on the first argument and linear on the second), such that $\langle v,v\rangle\in\Bbb R_{\ge0}$ for all $v\in V$.

Given two vectors $v\in V$ and $w\in V$, we shall prove that $|\langle v,w\rangle|\le\|v\|\|w\|$ where $\|\cdot\| = \sqrt{\langle\cdot,\cdot\rangle}$.

We can assume wlog that $$\lambda:=\langle v,w\rangle\in\Bbb R$$ (if $\lambda\notin\Bbb R$, simply replace $w$ by $\frac{|\lambda|}\lambda w$, which won't change the two members of the inequality we want to prove).

We shall use that $$\forall t\in\Bbb R\quad P(t):=\|v+tw\|^2\ge0$$

  • If $\|w\|=0$, this implies $2t\lambda+\|v\|^2\ge0$, so $$\forall s>0\quad|\lambda|\le\frac{\|v\|^2}{2s}$$ hence $\lambda=0$ and we are done.
  • If $\|w\|\ne0$, $$0\le P(-\lambda/\|w\|^2)=\|v\|^2-\lambda^2/\|w\|^2$$ hence $\lambda^2\le\|v\|^2\|w\|^2$ and again, we are done.

For the case of equality, you need to assume positive-definiteness. Essentially: if $\|w\|\ne0$ and $|\lambda|=\|v\|\|w\|$ then $P(-\lambda/\|w\|^2)=0$ implies $v-\frac\lambda{\|w\|^2}w=0$.

Anne Bauval
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    So basically the key is that C-S comes from P(t) >= 0 for all t. Clear with all cases accounted for, thanks for the detailed answer! – Tim Sep 28 '24 at 16:10