Q:

Using Linearity of Matrix Transposition.

Is the Following Proof Correct?
Theorem. Given that $\beta = \{A_1,A_2,A_3,...,A_n\}$ is a linearly independent subset of $\mathcal{M}_{n\times n}(\mathbf{F})$, prove that $\alpha = 
\{A_1^t,A_2^t,A_3^t,...,A_n^t\}$ is also a linearly independent set.
Proof. Assume that for some scalars $x_1,x_2,x_3,...,x_n\in\mathbf{F}$ we have $$0 = \sum_{j=1}^{n}x_jA_j^t\tag{1}$$
Now define the transformation $\mathcal{T}:\mathcal{M}_{n\times n}(\mathbf{F})\to \mathcal{M}_{n\times n}(\mathbf{F})$ such that $\mathcal{T}(A) = A^t$ where $A^t$ denotes the transpose of $A$, it is evident that $\mathcal{T}$ is linear.
Applying $\mathcal{T}$ to equation $(1)$ we see that 
$$0=\mathcal{T}(0) = \mathcal{T}\left(\sum_{j=1}^{n}x_jA_j^t\right) = \sum_{j=1}^{n}x_j\mathcal{T}(A_j^t) = \sum_{j=1}^{n}x_jA\tag{2}$$
the linear independence of $\beta$ implies that $x_1 = x_2 =x_3 =... = x_n = 0$ consequently $\alpha$ is linearly independent.

A:

It is correct, but you could have said the same thing with much less words, by noting that$$\sum_{j=1}^nx_j{A_j}^t=\left(\sum_{j=1}^nx_jA_j\right)^t.$$This is what you actually do, but there is no need to introduce the function $\mathcal T$.

