Is there a Linear Map to represent transpose?

Question

For example, I want $AB = A^T$, is there such a matrix $B$ that does this? Where $A$ is $n \times m$ and $B$ is $m \times n$.

Answer 1

If $A$ is $m\times n$ and $B$ is $n\times m$, then $AB$ is $m\times m$. On the other hand, $A^T$ is $n\times m$. Therefore one immediately sees that for $m\ne n$, there cannot be a $B$ with that property.

Therefore let's in the following assume $m=n$.

We have $$A = (A^T)^T = (AB)^T = (AB)B = AB^2$$ for all $A$, which is only possible if $B^2=I$, where $I$ is the $n\times n$ identity matrix.

On the other hand, since $B$ is supposed to transpose all $n\times n$ matrices, but $B$ itself is an $n\times n$ matrix, we also have $$B^T = BB = I.$$ In other words, $B$ is the identity matrix. But that means that $$A^T = AB = AI = A$$ for all $n\times n$ matrices $A$. However this is only true if $n=1$.

Therefore apart from the trivial case of $1\times 1$ matrices, there's no matrix $B$ with that property.

However, if you consider the matrix space itself as vector space, then the transposition itself is indeed a linear map, since $(\alpha A + \beta B)^T = \alpha A^T + \beta B^T$. However you cannot write that map as a single matrix. But if you accept more complicated expressions, you can indeed explicitly write down the transpose map:

Assume $A$ is an $m\times n$ matrix. Be $E_{ij}$ the $n\times m$ matrix which has an entry $1$ in line $i$ and column $j$, and all other entries $0$. Then you can write $$A^T = \sum_{i,j} E_{ij} \operatorname{tr} \left(E_{ij} A\right)$$ where $\operatorname{tr} X$ is the trace of $X$, that is, the sum of all diagonal elements.

This works as follows: Multiplying $A$ (which I'll assume to have matrix elements $a_{i,j}$) from the left with $E_{ij}$ gives an $n\times n$ matrix that has the $j$-th row of $A$ in the $i$-th row, and $0$ in all other rows. Therefore the only non-zero diagonal element is the one in row/column $i$, which contains the element $a_{ji}$. Multiplying that with $E_{ij}$ gives a $n\times m$ matrix that has $a_{ji}$ in row $i$, column $j$ and is $0$ otherwise. Adding over $i$ and $j$ finally gets you a matrix that for every $i$ and $j$ contains $a_{ji}$ in row $i$, column $j$, that is, $A^T$.

Answer 2

There can be no such $B$ that does this for all $A$. For example, if $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

then

$$ AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$

is impossible for any $B$.