A positive semidefinite bilinear form must be degenerate? If $\beta$ is a positive semidefinite (symmetric bilinear)/Hermitian form on $V$, i.e $\beta(v,v) \geq 0$ for all $v \in V$ with possibility that $\beta(v,v) = 0$ for $v \neq 0$. Then it satisfes the Cauchy-Schwarz inequality $$|\beta(v, w)|^2 \leq  \beta(v, v) \cdot \beta(w, w)$$
But if that's true, i.e there exists $v \neq 0$ with $\beta(v, v) = 0$ then $\beta(v, w) = 0$ for any $w \in V$ which implies that the  map $v \mapsto \beta(\cdot, v)$ such that $$\beta(\cdot, v) : V \to \mathbb{C} \quad w \mapsto \beta(v, w)$$
doesn't define an isomorphism  $V \cong V^{*}$, i.e there exists $v \neq 0$ such that $\beta(\cdot, v) = 0$ so $\beta$ is a degenerate form.
Is every positive semidefinite degenerate? That doesn't seem true, but I can't find my mistake.
 A: You are correct that a semi-definite, but not definite, inner product space $V$ is degenerate, in the sense that there is non-zero $v$ with $\langle v,w\rangle=0$ for every $w\in V$.
The proof that I know for Cauchy-Schwarz-Bunyakowski does not quite treat the semi-definite case, but easily adapts. Over $\mathbb R$ for simplicity, if $\langle w,w\rangle\not=0$, then $0\le \langle v+tw,v+tw\rangle$, and that quadratic function of $t$ takes its unique minimum at $\langle v,w\rangle/\langle w,w\rangle$... fed back into the inequality, gives $\langle v,w\rangle^2\le \langle v,v\rangle\cdot \langle w,w\rangle$. The obvious symmetric argument applies for $\langle v,v\rangle\not=0$.
Thus, $\langle w,w\rangle=0$ implies $\langle v,w\rangle=0$ for all $v$, and obviously with roles reversed.
Even more directly, when $\langle v,v\rangle = 0 = \langle w,w\rangle$, the inequality degenerates to
$0\le t\langle v,w\rangle$ for all real $t$... which also gives $\langle v,w\rangle=0$.
(This kind of conclusion also follows from "inertia theorems" for real-symmetric and complex-hermitian forms... )
