Can you transpose a matrix using matrix multiplication?

Say you have a matrix A = \begin{bmatrix}a&b\\c&d\end{bmatrix}

and I want it to look like $A^T$ = \begin{bmatrix}c&a\\d&b\end{bmatrix} Can this be done via matrix multiplication? Something like a matrix T such that $T*A = A^T$.

Solution 1:

If there were such a $T$, we would have that $T = T \times I = I^T = I$, where $I$ is the identity matrix. But then it would follow that $A = I \times A = T \times A = A^T$ for all matrices $A$; i.e., that all matrices are their own transposes. As this is not true, we conclude there cannot be any such $T$ as desired.

Solution 2:

Why not ? The answer is implicitly given by il maestro @ Qiaochu Yuan. Behind this, there is a tensor; in fact, the transposition is simply a permutation. We stack the considered matrix $A$, row by row, in a column: $\bar{A}=[a,b,c,d]^T$ and $\bar{A^T}=[a,c,b,d]^T$. Thus the tensor $T$ s.t. $T\bar{A}=\bar{A^T}$ is $\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$. Of course, we can generalize for every $n$.

We can congratulate yujinred for his ability to calculate the transpose of a $2\times 2$ matrix.

Solution 3:

You can try to find it!

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} X & Y \\ Z & T \end{pmatrix} = \begin{pmatrix} a & c \\ b & d \end{pmatrix} $$

Hence solve the associate system system:

$$\begin{cases} aX + bZ = a \\ aY + bT = c\\ cX + dZ = b\\ cY + dT = d \end{cases} $$

To find $X, Y, Z, T$.