Is matrix transpose a linear transformation? This was the question posed to me. Does there exist a matrix $A$ for which $AM$ = $M^T$ for every $M$. The answer to this is obviously no as I can vary the dimension of $M$. But now this lead me to think , if I take , lets say only $2\times2$ matrix  into consideration. Now for a matrix $M$, $A=M^TM^{-1}$ so $A$ is not fixed and depends on $M$, but the operation follows all conditions of a linear transformation and I had read that any linear transformation can be represented as a matrix. So is the last statement wrong or my argument flawed?
The operation that transposes &quot;all&quot; matrices is, itself, not a linear transformation, because linear transformations are only defined on vector spaces.
Also, I do not understand what the matrix $A=M^TM^{-1}$ is supposed to be, especially since $M$ need not be invertible. Your understanding here seems to be lacking...
However:
The operation $\mathcal T_n: \mathbb R^{n\times n}\to\mathbb R^{n\times n}$, defined by
$$\mathcal T_n: A\mapsto A^T$$
is a linear transformation. However, it is an operation that maps a $n^2$ dimensional space into itself, meaning that the matrix representing it will have $n^2$ columns and $n^2$ rows!
Notice that the answer from Troy Woo can be written simply as $(M^\top)_{ij}=\sum_{k,l=1}^{n}\delta^k_j\delta^l_im_{kl}$. Writing $I=(e₁,\ldots,e_n)$, then $M^\top=\sum_{i,j=1}^{n}e_ie_j^\top Me_ie_j^\top$, which is another form of the answer from Kevin Bumgartner.