Let $\varphi$ be an operator on a $k$-vector space $V$ with an inner product $\langle\cdot,\cdot\rangle$. Suppose that $\langle v,\varphi v\rangle = 0$ for every $v\in V$. If we take $k=\mathbb R$, is $\varphi$ is an isomorphism? What about if $k=\mathbb C$?
I think that both claims are false (in more generality). Can I get feedback on my proof and possibly alternate solutions (what happens in the infinite dimensional case)?
Suppose that $V$ is finite dimensional. Then $\varphi$ has an adjoint $\varphi^\ast$. Then we obtain $$ \langle \varphi^\ast v, v \rangle = \langle v,\varphi v \rangle = 0 $$ for every $v\in V$. Hence $\varphi^\ast v = 0$ for each $v$, which implies that $\varphi^\ast = 0$ and thus also that $\varphi^\ast(V)=\{0\}$. Since $\varphi^\ast(V)=(\ker\varphi)^\perp$, we obtain $\ker\varphi= V$. Thus $\varphi$ is not an isomorphism.
If $k=\mathbb R,$ then rotation by $90$ degrees is a nonzero operator which sends every vector to a vector orthogonal to the original one.
If $k=\mathbb C,$ then it is indeed impossible, but not for the reasons you stated. It actually follows from some polarization identities, which write $\langle Tu, w\rangle$ in terms of inner products of the form $\langle Tv, v\rangle.$ Since all of the latter types of inner products are $0,$ all of the former type are also $0,$ from which you can conclude that $T=0.$