# find the eigenvalues of a linear operator T

Let $A$ be $m*m$ and B be $n*n$ complex matrices, and consider the linear operator $T$ on the space $C^{m*n}$ of all $m*n$ complex matrices defined by $T(M) = AMB$.

-Show how to construct an eigenvector for $T$ out of a pair of column vectors $X, Y$, where $X$ is an eigenvector for $A$ and $Y$ is an eigenvector for $B^t$.

-Determine the eigenvalues of $T$ in terms of those of $A$ and $B$

-Determine the trace of this operator

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Suppose $u_A$ is a right eigenvector of $A$ corresponding to an eigenvalue $\lambda_A$. Similarly, let $u_B^T$ be a left eigenvector of $B$ corresponding to an eigenvalue $\lambda_B$. Then by a simple calculation, we have $T(u_A u_B^T) = \lambda_A \lambda_B u_A u_B^T$, hence $u_A u_B^T$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda_A \lambda_B$.

The trace can be calculated by summing the eigenvalues of $T$. This involves computing a formula assuming that $A,B$ are diagonalizable, and then using the fact that the diagonalizable matrices are dense coupled with continuity of $\mathbb{tr}$.

Alternatively, we can pick a basis for $\mathbb{C}^{m \times n}$ and compute the trace directly. A simple basis is given by $E_{ij} = e_i e_j^T$. If $X \in \mathbb{C}^{m \times n}$, let $[X]_{ij}$ denote the component of $X$ along $E_{ij}$, ie, $X = \sum_{i,j} [X]_{ij} E_{ij}$. Then the trace of $T$ is given by $\mathbb{tr}(T) = \sum_{i,j} [T(E_{ij})]_{ij}$. A simple computation shows that $[T(E_{ij})]_{ij} = [A]_{ii}[B]_{jj}$, from which the following formula follows: $$\mathbb{tr}(T) = \sum_{i,j} [A]_{ii}[B]_{jj} = \sum_i [A]_{ii} \sum_j [B]_{jj} = \mathbb{tr}(A) \mathbb{tr}(B).$$

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There is a standard way to do this kind of exercise. Firstly assume that $A$ and $B$ are diagonal. Then a short calculation shows that the eigenvalues are as given in previous solutions, i.e., the pairwise products of those of these matrices. The result then holds for diagonalisable matrices by a suitable choice of bases. The easiest way to get the final version is to use the fact that the diagonalisable matrices are dense in all matrices and employ a continuity argument involving the characteristic polynomials.

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Suppose $X$ is an e-vector for $A$ then $AX = \lambda_A X$ for some $\lambda_A \in \mathbb{C}$. Likewise, suppose $Y$ is an e-vector for $B^t$ then $B^tY = \lambda_B Y$ for $\lambda_B \in \mathbb{C}$. The linear operator $T: \mathbb{C}^{m \times n} \rightarrow \mathbb{C}^{m \times n}$ maps matrices to matrices. Given the data $X$ an $m \times 1$ vector and $Y$ a $n \times 1$ vector we need to construct an $m \times n$ object. The most natural guess is simply $XY^t$ which has the right size to be an "$M$" of the problem statement.

$$T(XY^t) = AXY^tB = AX(B^tY)^t$$

using the socks-shoes property of the transpose. Continuing,

$$T(XY^t) = \lambda_A X (\lambda_B Y)^t = \lambda_A\lambda_B XY^t.$$

this shows $T$ has e-vector $XY^t$ with e-value $\lambda_A\lambda_B$.

To calculate trace, you should think about what theorems are known connecting the trace and eigenvalues for an operator.

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You have found some of the eigenvectors and eigenvalues of $T$, but have you found all of them? –  Gerry Myerson Sep 4 '12 at 2:24
maybe not... I certainly didn't derive them. This is the trouble with guessing :) –  James S. Cook Sep 4 '12 at 2:48