Edit: andinos clarified to say he wants to know about the implicit mapping of the kernel function. Well I have bad news: It does not exist!! The proof works by showing there exist matrices $A,B$ such that the corresponding kernel matrix is not positive semi-definite. To finish, apply Mercer's theorem.
In particular, set $A = \left(\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right)$ and $B = A^T = \left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)$. Therefore $\textrm{tr}(AB) = \textrm{tr}(AA^T) = 4$, and $\textrm{tr}(BA)$ is identical. On the other hand, $\textrm{tr}(AA) = \textrm{tr}(BB) = 0$. therefore, the kernel matrix $K$ is $\left(\begin{array}{cc}0 & 4 \\ 4 & 0\end{array}\right)$. Set $x = \left(\begin{array}{c} 1 \\ -1\end{array}\right)$, and observe that $x^T K x = -8 < 0$, and therefore $K$ is not PSD, so the kernel $k(A,B) = \textrm{tr}(AB)$ is not PSD.
On the other hand! If you had instead defined your kernel to be $k'(A,B) = \textrm{tr}(AB^T)$, notice that $k'(A,B) = \sum_{i,j}A_{ij}B_{ij} = \Phi(A)^T\Phi(B)$ where $\Phi$ simply takes its input matrix and outputs it as a column vector.
