Finding the range of a 3x2 matrix

Question

How would you go about finding the range of a matrix like:

$\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 1 \\ \end{bmatrix}$

This one is confusing me because it maps to the third dimension while only having two column vectors, I'm thinking its a plane but how would you explicitly state what the range is?

Answer 1

Note that matrix maps from $\mathbb{R}^2$ to $\mathbb{R}^3$.

To determine the range, by RREF or by determinants, we can find that the matrix has rank $2$ and then the range is a subspace of dimension $2$ of $\mathbb{R}^3$ that is the plane

$$\pi:s\begin{bmatrix} 1 \\ 2 \\ 0 \\ \end{bmatrix}+t\begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}$$

Answer 2

Setting $$A=\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 1 \\ \end{bmatrix}\in\mathbb{R}^{3\times 2},$$ we have $\operatorname{rank}(A)=2$ if and only if there exists a $2\times 2$ submatrix of $A$ with determinant unequal to zero. We have 3 candidates here:

1. first and second row
2. first and third row
3. second and third row.

While any of these candidates work, easiest is candidate 2: the identity matrix with determinant 1. Thus, we have $\operatorname{rank}(A)=2.$