$\mathbb{R}^2, \mathbb{R}^3$ matrix/canonical basis

Question

let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear map and given by $f\pmatrix{1\\2}=\pmatrix{2\\1\\1}, f\pmatrix{2\\2}=\pmatrix{0\\4\\2}$

(1) Define the matrix of $f$ in regards to the canonical basis, meaning that the canonical basis is chosen in $\mathbb{R}^2$ and $\mathbb{R}^3$.

(2) Calculate the rank of this matrix.

My ideas for 1). I have calculated the image of the map for the canonical basis of $\mathbb{R}^2$:

$f\pmatrix{0\\1}= f\left(-\frac{1}{2} \pmatrix{2\\2}+\pmatrix{1\\2}\right)=\pmatrix{2\\-1\\0}$

$f\pmatrix{1\\0}= f\left(\pmatrix{2\\2}-\pmatrix{1\\2}\right)=\pmatrix{-2\\3\\1}$

so the first matrix is given by $M_1 = \pmatrix{-2 & 2\\3 & 1\\1 & 0}$. But I am stuck here, how do I proceed?

Answer 1

Your answer for the first part is correct (except for a missing $-$ for the first vector).

For the second part, the answer will probably depend a bit on your definition of the rank of a matrix. A standard way for determining the rank is to use Gaussian elimination in order to determine the row echelon form of the given matrix. Alternatively, you could think about what the maximal rank of a $3 \times 2$ matrix can be and then maybe find a reason why this actually is the rank here.