Linear transformation between matrices of different dimensions In my self-study of linear algebra, I have come across a problem that I can't really figure out. I have a linear transformation $f: \mathbb{R}^{2\times 2}\rightarrow\mathbb{R}^2$ with a linear transformation matrix in $\mathbb{R}^{2\times 4}$. I have to solve $f(x)=(1,1)$, but it is not clear how I would transform a $2\times 2$ matrix into a 2-vector by multiplying it by a $2\times 4$ matrix.
 A: In general, if a transformation matrix is $r \times s$, it gives you a map from $\mathbb{R}^s$ to $\mathbb{R}^r$. In your case then, the matrix defines you something from $\mathbb{R}^4$ to $\mathbb{R}^2$, but you want something from $\mathbb{R}^{2 \times 2}$ to $\mathbb{R}^2$. These things are the same though, if you have some way of identifying $\mathbb{R}^{2 \times 2}$ and $\mathbb{R}^4$ in a linear fashion, which should make sense since the two spaces have the same dimension. I suspect that a way of doing so is given in the context of the exercise? Such an identification could be as simple as
$$
  \begin{pmatrix} a & b \\ c & d \end{pmatrix} \to \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}.
$$
A: $\mathbb R^{2\times2}$ is a vector space over $\mathbb R$ of dimension $4$. So $\mathbb R^{2\times2} \simeq \mathbb R^4$. In order to choose a specific isomorphism between $\mathbb R^{2\times2}$ and $\mathbb R^4$, just pick a basis for $\mathbb R^{2\times 2}$. One possible basis is $E_{11},\ E_{12},\ E_{21},\ E_{22}$ where $E_{ij}$ is the $2\times 2$-matrix with $1$ in row $i$ and column $j$ and $0$ elswhere. Please convince yourself that this is indeed a basis of $\mathbb R^{2\times 2}$. With this basis, we have the following correspondence between $\mathbb R^{2\times 2}$ and $\mathbb R^4$.
$$
\mathbb R^{2\times 2} \ni \begin{pmatrix} a & b \\ c & d \end{pmatrix} \longleftrightarrow aE_{11} + bE_{12} + cE_{21} + dE_{22} \longleftrightarrow (a,b,c,d) \in \mathbb R^4$$
with $a,b,c,d \in \mathbb R$.

Now that we have a representation of $\mathbb R^{2\times 2}$ as row vectors in $\mathbb R^4$, you can write $f$ as a $2\times 4$-matrix and solve the equation $f(x) = (1,1)$ which will produce a certain row vector $x$. Using the above correspondence, you translate this row vector $x$ back into a $2\times 2$-matrix, and there you are.
