A linear transformation $f: \mathbb{R}^2 \to \mathbb{R}^3$ is defined by the following matrix: $$A (f;B,B')= 
 \begin{bmatrix}
 2 & -1\\
 1 & 3 \\
 -1 & 1 \\
 \end{bmatrix}
$$
$$B = ((1,3),(1,2)) $$
$$B′ = ((0,0,1),(1,0,−1),(0,1,0))$$
I want to find a basis for the kernel and image of the transformation. By solving the matrix to equal zero I found that $\operatorname\*{Ker}(f) = 0\_R$, so $\dim (\operatorname\*{Ker} (f))= 0$. This way I know the dimension of $\operatorname\*{Im}(f) = 2$. Is the basis of the image $\left<(2,1,-1),(-1,3,1)\right>$? Is the basis of the image **always** the linearly independent vectors of the matrix?