Definition
Let be $V$ and $U$ real vector spaces equipped with an inner product. So given a linear transformation $f:V\rightarrow U$ a function $f^*:U\rightarrow V$ is called the adjoint of $f$ if $$ \langle\vec u,f(\vec v)\rangle=\langle f^*(\vec u),\vec v\rangle $$ for all $\vec v\in V$ and for all $\vec u\in U$. In particular an endomorphism $f\in\mathscr L(V,V)$ is called Hermitian if $f=f^*$ and skew-Hermitian if $f=-f^*$.
Clearly if $f$ is skew-Hermitian then $$ \langle v,f(\vec v)\rangle=\langle f^*(\vec v),v\rangle=-\langle f(\vec v),v\rangle=-\langle\vec v,f(\vec v)\rangle $$ for all $\vec v\in V$ and so $\langle v,f(\vec v)\rangle=0$ for all $v\in V$. But unfortunately I don't be able to proce that if $f$ is such that $\langle v,f(\vec v)\rangle=0$ for all $v\in V$ then it is skew-Hermitian. So could someone hep me, please?
A common trick is to select two vectors $u$ and $v$ and apply the hypothesis to $u+v$. Using linearity you have $$0 = \langle u+v,f(u+v) \rangle = \langle u,f(u) \rangle + \langle u,f(v) \rangle + \langle v,f(u) \rangle + \langle v,f(v) \rangle = \langle u,f(v) \rangle + \langle v,f(u) \rangle$$ so that $$\langle u,f(v) \rangle = \langle -f(u),v\rangle.$$