Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

up vote 3 down vote accepted

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: $$\mathbb{tr}(T) = \sum_{i,j} [A]_{ii}[B]_{jj} = \sum_i [A]_{ii} \sum_j [B]_{jj} = \mathbb{tr}(A) \mathbb{tr}(B). $$

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.