Let $A$ be a general $n\times n$ invertible matrix. Let $T^A$ be the "transposer" matrix i.e. $T^A A = A'$. (Does that $T^A$ multiplied by $A$ equal the transpose of $A$?) Then does $T^A$ depend on the matrix $A$: is $T^A = T^B$ for all invertible Matrices? Prove your claim.
So can you take a matrix times the given matrix in order to find its transpose? If so, is that a general form that can be used for all invertible matrices?
I'm guessing it does not, but I am not totally sure or know how to go about proving that and I cannot find anything online about it.
Best Answer
There is no general transposer matrix.. For this, e.g. note that the only transposer for the identity in any dimension is the identity as the equation
$$T^II=I$$
needs to be fulfilled, but $T^II=T^I$.
However, for e.g. for non-symmetric matrices $A$, the identity is definitely not the transposer, as $A^\top\neq A$, but $IA=A$.
Now, we may try to construct the transposer matrix by considering $X=(x_{ij})_{i,j\leq n}$ and $A=(a_{ij})_{i,j\leq n}$, and then solving the equation $XA=A^\top$ for $X$. I.e. by the laws of matrix multiplication, you have to have
$$a_{ij}=\sum_{k=1}^n a_{jk}x_{ik}$$
for all $i,j$. However, this system of linear equations is not always solvable (only if $A$ is not symmetric). For this, consider the following example:
Look at $A=\begin{pmatrix}0 &1\\0 &0\end{pmatrix}$, then
$$\begin{pmatrix}x &y\\z &w\end{pmatrix}\begin{pmatrix}0 &1\\0 &0\end{pmatrix}=\begin{pmatrix}0 &x\\0 &z\end{pmatrix}\neq\begin{pmatrix}0 &0\\1 &0\end{pmatrix}$$