Finding a matrix representation of the transpose transformation Define $T : M_{n×n}(\mathbb{R}) → M_{n×n}(\mathbb{R})$ by $T(A) := A^t$. 
I know this transformation is linear and just takes a matrix and spits out it's transpose. I also know that the transpose is just a matrix with it's columns and rows swapped; however, I don't know how to form a matrix representation of this transformation for arbitrary $n$. 
Any help to get me started would be appreciated!
 A: $\newcommand{\R}{\mathbb R}$
Just write what you would do for any linear transformation: write a basis for $M(n,\R)$ then find the "effect" of the transformation on its elements.
In this case you'll formally have to go through an isomorphism between $M(n,\R)$ and $\R^{n^2}$: you want to represent it as a matrix, so it has to multiply some vectors. These vectors will represent the matrix in a given basis.
I'll make an example just in $M(2,\R)$, for the sake of simplicity.
A general matrix in this space is
$$
M=
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}.
$$
Now, let's choose a basis for the vector space: the obvious (but not the best*) choice is the standard basis $\mathcal S=\{E_{11},E_{12},E_{21},E_{22}\}$ given by $(E_{\mu\nu})_{ij}=\delta_{i\mu}\delta_{j\nu}$, so
$$
S=\left\{
\begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0 \\ 1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0 \\ 0 & 1
\end{pmatrix}
\right\}.
$$
In this basis, the matrix $M$ can be written as $aE_{11}+bE_{12}+cE_{21}+dE_{22}$ so it's easily represented as the vector in $\R^{2^2}=\R^4$
$$
v=
\begin{pmatrix}
a \\ b \\ c \\ d
\end{pmatrix}.
$$
Now, the transposition (I'll call it $T$) acts on the elements of the basis $\mathcal S$ as follows:
$$
T(E_{11})=E_{11},\ T(E_{12})=E_{21},\ T(E_{21})=E_{12},\ T(E_{22})=E_{22}.
$$
Therefore the matrix associated with $T$ is just
$$
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}.
$$

* The calculations were simple anyway, but there's a nicer way to do it (according to me, of course); it just requires a bit of knowledge a priori.
The matrix vector space $M(n,\R)$ is a direct sum of the space $S(n,\R)$ of symmetric matrices and $A(n,\R)$ of antisymmetric ones:
$$
M(n,\R)=S(n,\R)\oplus A(n,\R)
$$
so a basis of $M(n,\R)$ can be found from these two subspaces.
Moreover, $\dim A(n,\R)=\frac12(n-1)n$ and $\dim S(n,\R)=\frac12n(n+1)$.
Now, if $x\in A(n,\R)$ then $T(x)=-x$, and if $y\in S(n,\R)$ then $T(y)=y$, and this is obviously true for the elements of their bases too.
So, we form a basis of $M(n,\R)$ consisting of $\frac12n(n+1)$ linearly independent elements from $S(n,\R)$ and $\frac12n(n-1)$ elements from $A(n,\R)$.
The matrix representing $T$ is diagonal in this basis, since symmetric and antisymmetric matrices are all "eigenvectors" of $T$ (and we have a basis of them).
So $M$ will have 1 as the first $\frac12n(n+1)$ elements on the diagonal, and $-1$ on the remaining $\frac12n(n-1)$ ones.
I like this method because it is already general for any order $n$ of the matrices, without doing any calculations.
A: Let $e_{ij}$ be the matrix with a value of $1$ at entry $(i,j)$ and zero elsewhere.  This is a basis for the space $M_{n \times n}$.  Then you can define your transpose operation on that space as follows:
$$T(e_{ij}) = e_{ji}$$
If you want to display this as a matrix you will need to come up with an arbitrary ordering of $\{e_{ij}\}$.  Then you can use the above definition to find out which entries are $0$ or $1$.
A: The transformation can as you said be written as a linear transformation, but in the vector representation of the matrix:
$$
T(\mathrm{vec}(\mathbf(A))) = \mathbf{P}\mathrm{vec}(\mathbf(A))
$$
where $\mathbf{P}\in\mathbb{R}^{n^2\times n^2}$.  The matrix $\mathbf{P}$ is a permutation matrix known as a stride permutation or a perfect shuffle matrix.
It's a bit more complicated for $\mathbf{A}\in\mathbb{R}^{m\times n}$, but for the $2\times 2$ case this matrix is
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}.
$$
In general the matrix can be written as $\mathbf{I}_m\boxtimes \mathbf{I}_n$ where $\boxtimes$ is the box-product that for $\mathbf{C}\in\mathbb{R}^{m_1\times n_1}$ and $\mathbf{D}\in\mathbb{R}^{m_2\times n_2}$ is defined by
$$
\mathbf{\mathbf{C}\boxtimes\mathbf{D}}_{(i-1)m_2+j,(k-1)n_1+l}=c_{il}d_{jk}.
$$
The box-product is very similar to the Kronecker product.
A: Let's consider the case $n = 2$. We have the linear transformation 
$$ \begin{align*}T: M_{2 \times 2}(\Bbb R) &\to M_{2 \times 2}( \Bbb R) \\ A &\mapsto A^{\mathrm T} \end{align*} \; $$
Now we can identify the vector space $M_{2 \times 2}(\Bbb R)$ with $\Bbb R^{4} = \Bbb R^{2 \cdot 2}$ by identifiying 
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in M_{2 \times 2}(\Bbb R) \quad \text{with} \quad \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \in \Bbb R^4 \; .$$
Now, we just have to find the images of the standard basis of $\Bbb R^4$: 
$$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \mapsto \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, , \quad 
\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \mapsto \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \, , \quad
\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \mapsto \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \, , \quad 
\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \mapsto \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \; .$$
So now we can represent the linear trasformation $T$ with the representation matrix 
$$ M_T := \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \; .$$
So for $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in M_{2 \times 2}(\Bbb R)$, we get the image of $A$ under $T$ by using our tranformation matrix $A$ by calculating 
$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} a \\ c \\ b \\ d \end{bmatrix} \; ,$$
and this vector is identified with $\begin{bmatrix} a & c \\ b & d \end{bmatrix} = A^{\mathrm T}$. 
You can generalize this method for an arbitrary $n \in \Bbb N$.
