Finding a basis for the linear transformation S(A)=A^T?

[Eclair_de_XII]

Homework Statement

"Find $S_\alpha$ where $S: M_{2×2}(ℝ)→M_{2×2}(ℝ)$ is defined by $S(A)=A^T$.

Homework Equations

$A^T=\begin{pmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{pmatrix}$

$\alpha= \{ {\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}} \}$

The Attempt at a Solution

I've found $S_\alpha$ before where the vectors can be expressed as columns, but i have no experience with finding the transformation matrices for actual matrices. Can anyone help me? This is as far as I got:

$S(A)=A^T=\begin{pmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{pmatrix}=a_{11}\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}+a_{12}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}+a_{21}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}+a_{22}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

How would I form this into a four-dimensional basis?

 

[andrewkirk]

Flatten each of the 2 x 2 matrices into a 4-vector by concatenating its columns, eg
$$flatten\left(A^T\right) = \langle a_{11},a_{12},a_{21},a_{22} \rangle$$
Then the flattened versions of the four matrices on the RHS of your last equation, suitably re-ordered based on the order of their coefficients in $flatten(A)$, become the columns of the 4 x 4 transformation matrix.

 

[Eclair_de_XII]

Okay, I solved it. Thanks for telling me about that function...

 

[StoneTemplePython]

Note: in math this function is typically called the

vec() operator and has interesting uses with Kronecker products.

in certain programming languages, you'll see it called flatten (e.g. numpy.flatten() in python, though you need to specify column order as it defaults to row ordering.)