We define $T(X) = AX$ with $A$ & $X$ square matrices of size $n$ with complex entries.
First, write down all the eigenvalues (with their respective algebraic multiplicities) of $T$. Use this to compute the trace & determinant of $A$.
I have proven that $A$ and $T$ have the same eigenvalues. My guess is that each eigenvalue of $A$ occurs with algebraic multiplicity $n$ times in order to give us a total sum of $n^2$, the dimension of the space. This is the part I'd like some help proving.
Possible idea : Compute $T$ for a basis which puts it into upper triangular form and then read off the diagonal entries for the eigenvalues.
Any help would be great.
Suppose that the basis $\mathcal B = \{v_1,\dots,v_n\}$ triangularizes $A$ (so that $[A]_{\mathcal B}$ is upper triangular), and take any basis $\{w_1,\dots,w_n\}$ of $\Bbb R^n$. Let $\mathcal B^*$ denote the basis of $\Bbb C^{n \times n}$ given by $$ \mathcal B^* = \{v_i w_j^T : 1 \leq i,j \leq n\} $$ where the tuples $(j,i)$ are taken in lexicographical order. The matrix of $T$ relative to $\mathcal B^*$ is given by the block-diagonal matrix $$ [T]_{\mathcal B^*} = \pmatrix{[A]_{\mathcal B} \\ & \ddots & \\ && [A]_{\mathcal B}} = I \otimes [A]_{\mathcal B} $$ where $I$ denotes the identity matrix and $\otimes$ denotes the Kroneker product.