Trace of operator $\alpha(X)=AXB$ in terms of $A$ and $B$

Question

I'm currently doing this question. Fix a $A$ be a $m \times m$ matrix and $B$ be a $n \times n$ matrix, and let $X$ be a $m \times n$ matrix. Define a map $\alpha:V \to V$, where $V$ is a vector space of $m \times n$ matrices given by $\alpha(X)=AXB$. Find the trace of this linear operator.

Here's what I tried so far. I know that if I consider $\beta(X)=AX$, I can take a basis going down the columns of the matrix. (ie. I can take $\mathcal B=\{E_{11}, E_{21} ... E_{n1}, E_{12}, E_{22}...E_{nn}\}$, where $E_{ij}$ refers to the matrix with $1$ in it's $i$th row and $j$-th column and zero everywhere else. Then because $A$ acts on each column of $X$ the same, I have that:

$$ [\beta]_{\mathcal B} = \left(\begin{array}{rrrr} A & \cdots & & 0 \\ \vdots& A & & \\ & & \ddots & \vdots \\ 0 & & \cdots & A \end{array}\right) $$ where each $A$ is a block matrix.

For $\gamma(X)=XB$, I know $B$ acts on each row the same. I can take the basis $\mathcal{Y}=\{E_{11}, E_{12} \cdots E_{1n} \cdots E_{nn}\}$, this time going down the rows of the matrix, to have that $$ [\gamma]_{\mathcal Y}=\left(\begin{array}{rrrr} B^T & \cdots & & 0 \\ \vdots& B^T & & \\ & & \ddots & \vdots \\ 0 & & \cdots & B^T \end{array}\right) $$ But after this I don't know how to continue. Any hints? I can't just compose them together because the bases are different.

Are you familiar with the cyclic property of trace, $\mbox{tr}(XYZ)=\mbox{tr}(YZX)$? — Brian Borchers, Apr 22 '20 at 14:04
Yes, I am. An answer regarding this property would be nice as well — koifish, Apr 22 '20 at 14:08
@Brian Borchers Wait, are you saying that I can consider the operator BA acting on X instead and take the basis along the columns instead? Giving me $ntr(BA)$? — koifish, Apr 22 '20 at 14:30

score 1 · Answer 1 · answered Apr 22 '20 at 19:52

By vectorization your function can be expressed as $\operatorname{vec}(\mathbf{A} \mathbf{X} \mathbf{B}) = \left(\mathbf{B}^{\top} \otimes \mathbf{A}\right) \operatorname{vec}(\mathbf{X})$, and $\operatorname{tr}(\mathbf{B}^{\top} \otimes \mathbf{A}) = \operatorname{tr}(\mathbf{A}) \operatorname{tr}(\mathbf{B})$

score 0 · Answer 2 · answered Apr 22 '20 at 14:05

0

The trace of this operator is the sum of its eigenvalues, which have been determined here:

linear operator $f(X) = AXB$

answered Apr 22 '20 at 14:05

Dietrich Burde

130,978

score 0 · Accepted Answer · answered Apr 22 '20 at 20:22

Using coordinates, the trace of a linear operator $L$ on a vector space with basis $\{v_1,v_2,\ldots,v_n\}$ is the sum of the $i$-th coordinate of $L(v_i)$ over all indices $i$.

In your case, this means $\operatorname{tr}(\alpha)$ is the sum of the $(i,j)$-th entry of $\alpha(E_{ij})$ over all $(i,j)$s. Given any matrix $M$, its $(i,j)$-th entry can be obtained as $e_i^TMe_j$. Therefore $$ \operatorname{tr}(\alpha) =\sum_{i,j}e_i^T\alpha(E_{ij})e_j =\sum_{i,j}e_i^TAe_ie_j^TBe_j =\left(\sum_ie_i^TAe_i\right)\left(\sum_je_j^TBe_j\right) =\operatorname{tr}(A)\operatorname{tr}(B). $$

Trace of operator $\alpha(X)=AXB$ in terms of $A$ and $B$

3 Answers3