Seeing if a matrix is hermitian

Question

Let $V$ be a vector space of finite dimension , and $f:V \rightarrow V $ a linear map such that:

$$\langle f(u) | v \rangle = \langle u | f(v) \rangle$$ $\forall u , v \in V $

I have to prove that , if $B$ is an ortonormal base of $V$ , then $M_B(f)$ is hermitian.

I don't know from where to start. Some tip would be helpful.

Answer 1

If $B=\{b_1,b_2,...,b_n\}$ then taking $u=b_i,v=b_j$ shows that $(i,j)$ element of the matrix is same as the conjugate of the $(j,i)$ element. Hence the matrix is Hermitian.