How to find the conjugate transformation when the matrix representing the transformation is unavailable?

Question

I have a transformation $T:M_{nxn}^R \rightarrow M_{nxn}^R$ which is defined as follows: $T(A)=A^t$.

Correct me if I'm wrong but I don't think it's possible finding the matrix representing the transformation (I tried multiplying a general matrix of size 2x2 and found that I cannot create a matrix that copies $\alpha_{12}$ to the location of $\alpha_{21}$).

I need to find what might be the conjugate transformation of the one described above in order to decide if that transformation is unitary.

And to put it more formally, that $T^*T=TT^*=I_n$

Answer 1

I don't think it's possible finding the matrix representing the transformation

It will be impossible before you pick a basis. There is no "the matrix representing a transformation," because there are many representations, all depending on basis.

I tried multiplying a general matrix of size 2x2 and found that I cannot create a matrix that copies $\alpha_{12}$ to the location of $\alpha_{21}$

Did you perhaps try to multiply by a $2\times 2$ matrix to achieve the transformation? That was destined not to work, of course. $M_2(F)$ is a four dimensional vector space, so a linear transformation like transposition would need a $4\times 4$ matrix to express linear transformations. In general it will take an $n^2\times n^2$ matrix.

Let's pick a basis

Let's pick the basis that lets us use the convention of writing an $n\times n$ matrix row by row from top to bottom. Doing that, $\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$ becomes $(a,b,c,d,e,f,g,h,i)$. I will also consider these as row vectors being multiplied by matrices on the right side.

And now for examples

Here are examples for $n=2$ and $n=3$ which you can generalize. The matrix for the transpose map on $M_2(\Bbb F)$ becomes

$T=\begin{bmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1\\ \end{bmatrix}$

For $M_3(F)$:

$T=\begin{bmatrix} 1&0&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0&0\\ 0&0&0&0&0&0&1&0&0\\ 0&1&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0&1&0\\ 0&0&1&0&0&0&0&0&0\\ 0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0&1\\ \end{bmatrix}$

If you look at the matrix in $n\times n$ blocks, you can see the pattern emerging.

$T^*T=TT^*=I_n$

Yep. If you're working in $\Bbb C$, then you see these matrices are, in fact, already symmetric and satisfy $T^\ast=T$. You'll find that $T^2=I_n$, as one would expect with the transpose map.

Answer 2

Another approach is to think in terms of operators rather than matrices. This linear operator preserves norms, therefore it's unitary.

We could also show that $T$ preserves inner products. The inner product of matrices $A, B \in M_{n \times n}(\mathbb R)$ is \begin{equation} \langle A, B \rangle = \text{trace}(A^T B). \end{equation} Thus $\langle T(A), T(B) \rangle = \text{trace}(A B^T) = \text{trace}(B^T A) = \langle B, A \rangle = \langle A, B\rangle$.

Here's another argument. We will find the adjoint of $T$. $\langle A, T(B) \rangle = \text{trace}(A^T T(B)) = \text{trace}(A^T B^T) = \text{trace}(BA) = \text{trace}(AB) = \langle T(A), B \rangle.$ This shows that $T^* = T$. It follows that $T T^* = T^* T = T^2 = I$.

(I used the fact that $\text{trace}(AB) = \text{trace}(BA)$.)