Confused about dimension of a range of a linear map Suppose you have a linear map $T:Mat_{2\times3}(R)\rightarrow Mat_{2\times2}(R)$
Why is the range(T) 4 at most? I thought that this is a 2x2 matrix and that the maximum range is 2?
 A: 
Why is the [dimension of] range(T) 4 at most? I thought that this is a 2x2 matrix and that the maximum range is 2?

$\text{range}(T)$ is not a $2\times 2$ matrix, but a SET of $2\times 2$ matrices. Precisely, it is a subspace of $W=M_{2\times 2}(\mathbb R)$. Its dimension has no connection with the maximum rank of the elements of $W$.
The dimension of the range of a linear application $f: V\to W$ is at most the minimum between the dimensions of $V$ and $W$. Since $\dim(M_{n\times m}(\mathbb R))=nm$, the dimension of $T$ is at most $2\times 2=4$.
If you choose a basis of $V=M_{2\times 3}(\mathbb R)$ and a basis $M_{2\times 2}(\mathbb R)$, then you can represent $T$ by a $4\times 6$ matrix (whose rank will be the dimension of $\text{range}(T)$).
A: Range$(T)$ is a subspace of $M_{2\times 2}(R)$, which is a vector space of dimension $4$. Because the dimension of any subspace is less than or equal to its parent space, it follows that the largest possible dimension of Range$(T)$ is $4$.
You can verify that the dimension of $M_{2\times 2}(R)$ is $4$ by verifying that the list of matrices $$\begin{bmatrix}1 & 0 \\ 0& 0 \end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0& 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1& 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0& 1 \end{bmatrix} $$ is linearly independent and spans $M_{2\times 2}(R)$.
