How to prove Schwarz inequality for Hermitian forms?

Question

I'm trying to do something like the proof of the Schwarz inequality for inner product.

If $h(y,y)\neq 0$, then we can take $\alpha=-h(x,y)/h(y,y)$ and calculate $h(x+\alpha y,x+\alpha y)$ which is a nonnegative number. The desired conclusion follows of this calculation.

If $h(y,y)=0=h(x,y)$, then the inequality is trivial.

My question is: how do we know that the case $h(y,y)=0$ and $h(x,y)\neq 0$ is not possible? When we are working with an inner product, this case is not possible because, by definition, $\langle y,y\rangle=0\Rightarrow y=0\Rightarrow\langle x,y\rangle=0$. But we don't have this condition for hermitian forms.

Thanks.

Answer 1

If $h(y,y) = 0$ but $h(x,y) \ne 0$, then taking $\alpha = - t h(x,y)$ with $t > 0$ we would have $h(x + \alpha y, x + \alpha y) = h(x, x) - 2 t |h(x,y)|^2$, which would be negative for sufficiently large $t$.

Answer 2

The following proof is taken from S. Lang, Linear algebra.

Let $\alpha = h(w,w)$ and $\beta = -h(v,w)$. Then, writing $\|x\|^2 = h(x,x)$ for simplicity, $$ 0 \leq h(\alpha v + \beta w,\alpha v + \beta w) = \alpha \bar{\alpha} h(v,v) + \beta \bar{\alpha} h(w,v) + \alpha \bar{\beta} h(v,w) + \beta \bar{\beta} h( w,w). $$ Hence $$ 0 \leq \|w\|^4 \|v\|^2 - 2 \|w\|^2 h(v,w) \overline{h(v,w)} + \|w\|^2 h(v,w) \overline{h(v,w)} $$ and finally $$ \|w\|^2 |h(v,w)|^2 \leq \|w\|^4 \|v\|^2. $$ If $w=0$, the conclusion is trivial; otherwise we divide by $\|w\|^2$ and take a square root.