I was reading the definitions:
Let $X$ be a vector space and $f: X \times X \longrightarrow \mathbb K$, where $\mathbb K = \mathbb R$ or $\mathbb C$. $f$ is said to be a Hermitian form on $X$ if: $\forall$ $x,y,z \in X$ and $\lambda \in \mathbb K$:
$1)$ $f(x+y,z) = f(x,z) + f(y,z)$
$2)$ $f(\lambda x,y) = \lambda f(x,y)$
$3)$ $f(x,y) = f(y,x)^*$, where $^*$ indicates the complex conjugate.
If, in addition, $\forall$ $ x \in X$, $f(x,x) \ge 0$, then $f$ is said to be a non-negative Hermitian form on $X$.
$f$ is said to be an inner product on $X$ if it is a non-negative Hermitian form and $\forall$ $x$, $f(x,x) = 0 \implies x = 0$
I can think of some non-negative Hermitian forms and I already have many examples, but they all happen to be inner products as well.
What is an example of a non-negative Hermitian form which is still not an inner product?
That is, $\exists$ $x \in X \setminus \{0\}$ such that: $f(x,x) = 0$.
Thanks.
Comment promoted to answer, per request of OP.
Let the vector space be ${\bf R}^2$, and let $f((a,b),(c,d))=ac$. Note that $f((0,17),(0,17))=0$.