Basis of image of a linear transformation

Question

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?

Answer 1

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be a linear map. Suppose that $A\in \mathbb{R}^{m\times n}$ is the matrix of $f$ w.r.t. the standard bases. Since the rank of $A$ is equal to the column rank of $A$, it suffices to show that the columns of $A$ vectors in the image of $f$. Now let $e_i$ be the $i$-th standard basis vector of $\mathbb{R}^n$. Then $f(e_i)=Ae_i=A_i$ where $A_i$ is the $i$-th column of $A$. This completes the proof.

Can you figure what happens the matrix of $f$ is not given w.r.t. the standard bases.?

Edit: You edited the question. The matrix $A$ is no longer the matrix of $f$ w.r.t. the standard bases. The above argument fails since the columns of $A$ do not necessarily belong to the image. However, the columns represent the coordinates of images of basis vectors. So what does belong to your image?

Second Edit: Let's work with your example. You know that basis of $f$ w.r.t. the bases $B$ en $B'$. What information does give us? We know that $f((1,3))=2(0,0,1)+1(1,0,-1)-1(0,1,0)$. Notice that the coordinates appearing in this last expressions are $(2,1,-1)$ which is exactly the first column of $A$. If we want to write $(1,3)$ w.r.t. the basis $B$, then we bet $(1,3)=1(1,3)+0(1,2)$. Notice that $A\begin{pmatrix}1\\0 \end{pmatrix}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$. This result is the coordinate of $f((1,3))$ w.r.t. the basis $B'$. We do not have that $(2,1,-1)\in \text{Im}(f)$, but $f((1,3))=2(0,0,1)+1(1,0,-1)-1(0,1,0)=(1,-1,1)\in \text{Im}(f)$. In the same fashion $f((1,2))=-1(0,0,1)+3(1,0,-1)+1(0,1,0)=(3,1,-4)\in \text{Im}(f)$. Hence $\text{Im}(f)=\text{Span}\left\{(1,-1,1),(3,1,-4)\right\}$.

Answer 2

No, not always. The image can have many bases. But the converse is true. That is, the linearly independent collumns of a matrix are always a basis of the image.