$T: \mathbb{R}^n \to \mathbb{R}^m$ and relationship with dimensions of matrix

Question

If $T(x) = Ax$.

If $A$ is a $n \times m$ matrix. ($n$ = rows, $m$ = cols)

Is $T: \mathbb{R}^n \to \mathbb{R}^m$ always the values of dimensions of the matrix $n \times m$?

So in other words, the domain is the row space and codomain the column space?

Answer 1

Yes, exactly, $n\times m$ matrices with real entries correspond to $\Bbb R^m\to\Bbb R^n$ linear maps:

If $A$ is an $n\times m$ matrix, then it can be multiplied from the right by $m\times 1$ column vectors, and exactly this multiplication will be the action as linear map: $$\Bbb R^m\ni\ \ x\mapsto A\cdot x \ \ \in \Bbb R^n\,.$$ Conversely, if a linear map $T:\Bbb R^m\to\Bbb R^n$ is given, construct the matrix $$A:= \left[T(e_1) \ \ T(e_2) \ \ \dots \ \ T(e_m) \right]$$ using the standard basis $e_1,e_2,...e_m$ of $\Bbb R^m$. Then, by linearity, for any $x\in\Bbb R^m$ we can get that $T(x)=A\cdot x$.

Answer 2

Yep, that's close. The row space is contained in the domain and the column space is contained in the codomain. The relationship is pretty tight. A couple other things to keep in mind:

the columns of the matrix are the images of the standard basis vectors. So, for example, the first column of the matrix is the image of $\left(1,0,\ldots,0\right)$. The second column is the image of $\left(0,1,\ldots,0\right)$, etc.

The rows, which are elements of $\mathbb{R}^n$, also span a very interesting subspace that's a little harder to explain. Basically, the matrix is one-to-one on the subspace of $\mathbb{R}^n$ spanned by the rows, and the matrix gives an isomorphism between this space and the image. Another way to think about it, is that the span of the rows is the orthogonal complement to the kernel.

Alternatively, note that the rows are the columns of the transpose of the matrix, so they are the images of the standard basis vectors under the transformation given by the transpose.