Let $A$ be the matrix, $$A=\begin{bmatrix} 3 & 0 & 0 & 2\\ 0 & 3 & 2 & 0\\ 0 & 2 & 3 & 0\\ 2 & 0 & 0 & 3 \end{bmatrix}$$ And the linear transformation $T : \mathbb{R}^{4 \times 4} \rightarrow \mathbb{R}^{4 \times 4}$ by $T(X)=AX$ then what is the trace of $T$?
So I computed the $16 \times 16$ matrix representing $T$ and it turns out the trace is 48. But can we compute the trace just by looking at $A$?
If you choose the standard basis for $M_4(\mathbb{R})$ in the correct order, the matrix representing $T_A(X) = A \cdot X$ will be a $16 \times 16$ block matrix that consists of four copies of $A$. This works for all $A \in M_n(\mathbb{R})$ and hence
$$ \mathrm{trace}(T_A) = n \cdot \mathrm{trace}(A). $$
Similarly, you can relate the characteristic polynomial of $T_A$ to $A$, the eigenspace of $T_A$ to $A$ and so on.