Eigenvectors of a linear operator Let $A$ be $m \times m$ and $B$ be $n \times n$ complex matrices, and consider the linear operator $T$ on the space $\mathbb{C}^{m\times n}$ of all $m \times n$ complex matrices defined by $T(M)=AMB$.
(a) Show how to construct an eigenvalue for $T$ out of a pair of column vectors $X,Y$, where $X$ is an eigenvector for $A$ and $Y$ is an eigenvector for $B^t$.
What I've tried: I tried using the definitions: $X$ eigenvector of $A$: $X \neq 0$ and $( \exists \lambda \in \mathbb{C}, AX= \lambda X)$ and $Y$ eigenvector of $B^t$: $Y \neq 0$ and $( \exists \mu \in \mathbb{C}, B^tY= \mu Y)$
I tried combining the two in order to get an expression similar to $T(M)$: $AX(Y^tB)^t=\lambda\mu XY$ but it doesn't seem to lead be anywhere.
I haven't looked at the other questions, but I will still post them:
b) Determine the eigenvalues of $T$ in terms of those of $A$ and $B$.
c) Determine the trace of this operator.
 A: Suppose $Ax=\lambda x$, $B^* y = \mu y$. Then $T(x y^*) = Ax y^* B = (Ax) (B^*y)^* = (\lambda x) (\mu y)^* = \lambda \overline{\mu} x y^*$. Hence $x y^*$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda \overline{\mu}$.
(b) follows from this, but you will need to deal with the case when $A,B^*$ do not have a full set of eigenvectors. One way is rely on 'continuity' of eigenvalues and find 'nearby' $A_n,B_n$ that have a full set of eigenvectors. Some attention to detail is needed.
(c) follows from (b) using the fact that the trace is the sum of the eigenvalues.
Addendum: More detail for (b):
If $A,B^*$ both have distinct eigenvalues, then both have eigenvectors $u_1,...,u_m$ and $v_1,...,v_n$ that span $\mathbb{C}^m$ and $\mathbb{C}^n$ respectively. Then $u_i v_j^*$ form a basis for $\mathbb{C}^{m \times n}$. To see this, suppose $\sum \alpha_{ij} u_i v_j^* = 0$. 
Let $U,V$ be the corresponding matrices formed by stacking the columns together. Then we can write this as $\sum \alpha_{ij} u_i v_j^* = \sum \alpha_{ij} U (e_i e_j^T) V^* =  U(\sum \alpha_{ij} e_i e_j^T)V^* = 0$. Since both $U,V$ are non-singular, it follows that $\alpha_{ij} = 0$. In this basis, $T$ is diagonal, and the characteristic polynomial is given by $\chi_T(x) = \Pi_{\lambda \in \sigma(A), \mu \in \sigma(B^*)} (x-\lambda \overline{\mu}) =
\Pi_{\lambda \in \sigma(A), \mu \in \sigma(B)} (x-\lambda {\mu})$. 
One way to deal with repeated eigenvalues is to approximate $A,B$ with matrices $A_i,B_i$ that have distinct eigenvalues.
Let $\chi_T$ be the characteristic polynomial of $T$. Suppose $A_i \to A$ and $B_i \to B$,and let $T_i(X) = A_i X B_i$. Let $\chi_{T_i}$ be the characteristic polynomial of $T_i$.  The coefficients of the characteristic polynomial are continuous functions of the constituent matrix, hence we have $\chi_{T_i} \to \chi_T$ (the space $\mathbb{P}^{(nm+1)}$ of polynomials of degree $nm+1$ or less is finite dimensional, so we can use any suitable norm to characterize convergence).
To finish, we must find suitable $A_i,B_i$ and show that $\chi_{T_i}$ converges to a polynomial of the required form.
Let $\Delta_k=\operatorname{diag}(1,...,k)$ be a diagonal matrix. Let $U^{-1}AU =J_A$ be the Jordan form of $A$. Let $A_\epsilon = U(J_A+\epsilon \Delta_m)U^{-1}$.
We see that $\sigma(A_\epsilon) = \{ [J_A]_{ii}+i \epsilon\}_{i=1}^m$. Hence for some $\delta_A>0$, if $\epsilon \in (0,\delta_A)$, we have $|\sigma(A_\epsilon)| =m$.
A similar construction gives a $B_\epsilon$ such that for some
$\delta_B >0$, then $\sigma(B_\epsilon) = \{ [J_B]_{ii}+i \epsilon\}_{i=1}^n$ and $|\sigma(B_\epsilon)| =n$, for all $\epsilon \in (0,\delta_B)$.
Abusing notation slightly, we let $A_i = A_{\frac{1}{i}}, B_i = B_{\frac{1}{i}}$. Hence for sufficiently large $n$ we have 
$\chi_{T_i}(x)= \Pi_{a=1}^m \Pi_{b=1}^n (x-([J_A]_{aa}+\frac{a}{i}) {([J_B]_{bb}+\frac{b}{i})})$, and we see $\chi_{T_i} \to \xi$, where $\xi(x) = \Pi_{a=1}^m \Pi_{b=1}^n (x-[J_A]_{aa} {[J_B]_{bb}}) = \Pi_{\lambda \in \sigma(A)} \Pi_{\mu \in \sigma(B)} (x-\lambda  {\mu})^{\alpha_A(\lambda) \alpha_B(\mu)}$,
where $\alpha_A$ and $\alpha_B$ are the algebraic multiplicities of the respective eigenvalues.
Hence we have $X_T(x) = \Pi_{\lambda \in \sigma(A)} \Pi_{\mu \in \sigma(B)} (x-\lambda {\mu})^{\alpha_A(\lambda) \alpha_B(\mu)}$.
As an aside, note that the multiplicity of each eigenvalue $\nu = \lambda {\mu}$ is given by $\alpha_T(\nu) = \sum_{\substack{\lambda {\mu} = \nu \\ \lambda \in \sigma(A), \mu \in \sigma(B)}} \alpha_A(\lambda) \alpha_B(\mu)$.
(c) follows since the trace is the sum of the eigenvalues, hence
$\operatorname{tr}T = \sum_{\lambda \in \sigma(A)} \sum_{\mu \in \sigma(B)}  \alpha_A(\lambda) \alpha_B(\mu) \lambda {\mu}$.
A: @copper.hat, the complex numbers have nothing to do here; $T=A\otimes B^T$ is a Kronecker product, a $mn\times mn$ matrix, cf. 
http://en.wikipedia.org/wiki/Kronecker_product
and is defined over any field. The $mn$ eigenvalues of $T$ are the $(\lambda_i\mu_j)_{i,j}$.
In particular $trace(T)=trace(A)trace(B)$.
