Is it possible to transpose a square matrix by multiplication?

Question

Is it possible to transpose matrix by left and/or right multiplication?

$XAY = A^T$

Do $X$ and $Y$ exist for any $A$?

Answer 1

Yes, since every square matrix is similar to its transpose. Therefore, you can even take $Y=X^{-1}$.

Answer 2

What is certainly not possible is to find and $X$ and $Y$ that satisfies $X A Y = A^T$ for any $A$ (i.e., which does not depend on $A$). To see why, consider the following example. Let $A_{i,j} = e_i e_j^T$, where $e_i$ is a vector that is zero except for a single $1$ at position $i$. The matrix $A_{i,j}$ therefore contains a single one at index $(i,j)$. Let's try to find an $X$, $Y$:

$$X A_{i,j} Y = X e_i e_j^T Y = x_i y^j \stackrel{!}{=} A_{j,i}^T = e_j e_i^T,$$

where $x_i$ is the $i$-th row of $X$ and $y^j$ is the $j$-th row of $Y$. Now, the set of solutions to the above condition is given by $x_i = c \cdot e_j$, $y^j = \frac 1c \cdot e_i^T$ for any $c \in \mathbb{R}_{\neq 0}$. Therefore, for $A_{1,1}$ the first column of $X$ should be proportional to $e_1$ whereas for $A_{1,2}$, the first column of $X$ should be propostional to $e_2$, which contradicts. It is therefore not possible to find $X$ and $Y$ that transposes all $A_{i,j}$ (not even one pair, for that matter).

If you let it be dependent on $A$, there are several ways to find an $X$ and $Y$ even if $A$ is not invertible. For instance, let the singular value decomposition of $A$ be $A = U \Sigma V^T$. Since $A$ is quadratic, $U$ and $V$ are square orthonormal matrices. Then $A^T = V \Sigma U^T$ so the matrices $X = Y = V U^T$ (which are themselves unitary) should do the trick. Since any matrix has an SVD, this shows that it works even for singular and non-diagonalizable matrices.

Answer 3

Alt. Sol:

Suppose $X\in M_{m\times n}(\mathbb{R})$ is not symmetric and that there exists matrices $A$ and $B$ such that $AXB = X^T$.

Then we would have: $$(AXB)^T = (X^T)^T = X$$ $$\implies B^TX^TA^T = X$$ $$\implies B^T(AXB)A^T = X$$ $$\implies (B^TA)X(BA^T) = X$$ $$\implies (A^TB)^TX(BA^T) = X.$$ But this says essentially that: $$A^TB = Id_{m\times m}\text{ }\text{ }\text{ and }\text{ }\text{ }BA^T = Id_{n\times n}$$ In other words that $A^T$ is a left and a right inverse for $B$. So we may write $B^{-1}= A^T$.

Going back to the first equation then: $$AXB = X^T$$ $$\implies AX = X^TA^T = (AX)^T.$$ This says the matrix $AX$ is symmetric. Element wise, this gives:

$$(AX)_{ij} = \sum\limits_{k=1}^ma_{ik}x_{kj} = \sum\limits_{l=1}^ma_{jl}x_{li} = (AX)_{ji}$$

From this it can be seen that regardless of symmetry of $A$, since $X$ is not symmetric, if $i\neq j$ then some $x_{kj}$ must not have a match for any $x_{li}$. So the sums can't be the same and this is our contradiction.

We may conclude that for a given non-symmetric matrix, no such $A$ and $B$ exist yielding $AXB = X^T$. $\blacksquare$

---

Note: If $X$ is symmetric, then $A$ and $B$ can be chosen to be the identity matrix.