What is the matrix representation of the linear space isomorphism $\mathbb{R}^4\cong\mathbb{R}^{2\times2}$?

Question

Every linear map between finite dimensional vector spaces can be represented by a matrix, once bases are chosen for the domain and range spaces.

Question: What is the matrix representation for the linear space isomorphism \begin{align} \mathbb{R}^{2\times2}&\rightarrow\mathbb{R}^4\\ \begin{bmatrix}a&b\\c&d\end{bmatrix}&\mapsto[a\;b\;c\;d] \end{align} between $\mathbb{R}^4$ and the space $\mathbb{R}^{2\times2}$ of $2\times2$ matrices, given the standard bases of these spaces?

Edit: Let's say the standard basis of $\mathbb{R}^{2\times2}$ is: $$ \bigg\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\bigg\} $$

Answer 1

Given a map $V \to U$ with bases $\{e_i\}_{i=1}^{n}$ and $\{u_j\}_{i=1}^n$ respectively, a matrix form is given by forming columns $[u_1 \dots u_n]$ where $e_i \to a_{1i}u_1+ \dots a_{ni}u_n$.

In your case the map is $e_i \to u_i$, where $u_i$ is the standard basis for $\mathbb R^4$.

This means that the matrix representation is just the identity.

Answer 2

Using the (ordered) basis you provided for $\Bbb R^{2\times 2}$ and the standard basis for $\Bbb R^4$, your isomorphism is given by the identity matrix! Check it out. (By the way, most people will denote elements of $\Bbb R^4$ by column vectors.)

Answer 3

This is a short answer to my own question.

In the given basis of $\mathbb{R}^{2\times2}$, we have that $$ \begin{bmatrix}a&b\\c&d\end{bmatrix}=a\begin{bmatrix}1&0\\0&0\end{bmatrix}+b\begin{bmatrix}0&1\\0&0\end{bmatrix}+c\begin{bmatrix}0&0\\1&0\end{bmatrix}+d\begin{bmatrix}0&0\\0&1\end{bmatrix}, $$ so this matrix is just the point in the $4$-dimensional vector space $\mathbb{R}^{2\times2}$ with coordinates $$ \begin{bmatrix}a\\b\\c\\d\end{bmatrix}. $$ Hence, the desired linear space isomorphism is represented by the identity matrix.