$V=\mathbb{M}_{n}(\mathbb{F})$, $f\in V^{*}$ with $f(AB)=f(BA),$ exists $\lambda\in\mathbb{F}$ with $f(A)=\lambda\textrm{tr}(A)\,\forall \,A\in V$

Question

Let $V=\mathbb{M}_{n}(\mathbb{F})$ the space of $n\times n$ matrices. Let $f\in V^{*}$ satisfying $f(AB)=f(BA).,$ So, exists unique $\lambda\in\mathbb{F}$ such that $f(A)=\lambda\textrm{tr}(A)\,\forall \,A\in V$, where $\textrm{tr(A)}$ denotes the trace of $A\in V$.

Any tips to follow will help me. I think that $\lambda$ has something to do with the eigenvalues of $A$, but I don't know.