Let $V$ be a finite-dimensional vector space, and let $\langle\cdot,\cdot\rangle$ be a nondegenerate bilinear form on $V$. If $T$ is a linear operator of $V$, does it follow that $T$ and $T^*T$ have the same kernel?
This question addresses the case when $V$ is a f.d. vector space over $\Bbb R$ and $\langle\cdot,\cdot\rangle$ is positive definite. I can show that if there is some scalar $\alpha\in F$ such that $\alpha$ is not an eigenvalue of $T$, then the claim holds.
Suppose $\alpha$ is not an eigenvalue of $T$. Then $T-\alpha I$ is a nonsingular operator on $V$ and hence an isomorphism. Suppose that $v\in\ker T^*T$. Then for all $w\in V$, we have that $$0=\langle 0,w\rangle=\langle T^*Tv,w\rangle=\langle Tv,Tw\rangle=\langle T^*Tv,\alpha w\rangle$$ Thus $$0=\langle (T-T^*T)v,(T-\alpha I)w\rangle$$ for all $w\in V$. Since $\langle\cdot, \cdot\rangle$ is nondegenerate, we have that $T^*Tv=Tv=0$. Thus $\ker T^*T\subseteq\ker T$. The other inclusion is obvious.
This covers a lot of cases, but I cannot prove the claim in general nor construct a counter-example. Any help in either direction is appreciated.
What about $V =(\mathbb F_2)^2$ with $\langle u, v \rangle = u^T v$ and $T = T^* = \pmatrix{1 & 1\cr 1 & 1\cr}$? Then $T^* T = 0$. This shows, btw, that your proof is incorrect, because $1$ is not an eigenvalue of $T$.