Linear map and trace

Let $f:\mathbb{F}^{n \times n} \rightarrow \mathbb{F}$ be a linear map such that $f\left ( AB \right ) = f \left ( BA \right )$  forall $A, B \in \mathbb{F}^{n \times n}$. Prove that there exists a $\kappa \in \mathbb{F}$ such that $f\left ( A \right ) = \kappa \;{\rm tr} \left ( A \right )$ forall $A \in \mathbb{F}^{n \times n}$.

Solution

For $i, j \in \{1, 2, \dots, n \}$ let us consider the matrix ${\rm E}_{ij}$ that has $1$ as an entry at $(i, j)$ and zero everywhere else. Observe that:

$${\rm E}_{ij} \; {\rm E}_{k \ell}= \left\{\begin{matrix}
\mathbb{O} &,  &j \neq k \\
 E_{i \ell}& ,  & j =k 
\end{matrix}\right.$$

Thus for every $i, j \in \{1, 2, \dots, n\}$ such that $i \neq j$ we have that ${\rm E}_{ij} = {\rm E}_{i1} \; {\rm E}_{1j}$. Hence, using the assumption we get that

$$f\left ( {\rm E}_{ij} \right ) = f \left ( E_{i1} \; {\rm E}_{1j} \right ) = f \left ( {\rm E}_{1j} \; {\rm E}_{i1} \right ) = f\left ( \mathbb{O} \right ) = 0$$

Also , for every $i, j \in \{1, 2, \dots, n\}$ we have that

$$f \left ( {\rm E}_{ii} \right ) = f \left ( {\rm E}_{i1} \; {\rm E}_{1i} \right )= f \left ( {\rm E}_{1i} \; {\rm E}_{i1} \right ) = f \left ( {\rm E}_{11} \right )$$

Let $\mathcal{A} \in \mathbb{F}^{n \times n}$ and note that $\displaystyle \mathcal{A} = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} {\rm E}_{ij}$. Since $f$ is linear we have that:

\begin{align*}
f\left ( \mathcal{A} \right ) &=f \left ( \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} {\rm E}_{ij} \right ) \\
 &= \sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij} f \left ( {\rm E}_{ij} \right )\\
 &= \sum_{i=1}^{n} a_{ii} f \left ( {\rm E}_{ii} \right ) \\
 &= f \left ( {\rm E}_{11} \right ) \sum_{i=1}^{n} a_{ii} \\
 &= f \left ( {\rm E}_{11} \right ) \; {\rm tr} \; \left ( \mathcal{A} \right )
\end{align*}

The exercise can also be found at mathematica.gr