Basis for Kernel and image of map Consider $V= \left \{ v= \begin{bmatrix}
x_1\\ 
x_2\\ 
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by
$f\begin{pmatrix}
x_1\\ 
x_2\\ 
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\ 
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)
Find the basis for the kernel and the image. Find their dimension.
my thoughts
I have shown it is not a isomorphism, that $V$ is a subspace and - I think the kernel would be $\begin{pmatrix}
1 &-1  &2 \\ 
 1&0  &-1 \\ 
 0&  1& -3
\end{pmatrix}$ and after reducing to reduced echelon form I get that the basis would be $(1,3,1)$ with dimension $1$ because of the free variable $x_3$. Now, for the image, I have no idea since I do not know how to find the image in the first place to use Gaussian elimination.
 A: By definition$$\ker(f)=\left\{\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\in V\,\middle|\,f\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right)=\begin{bmatrix}0\\0\end{bmatrix}\right\}.$$So, you can compute it by solving the system$$\left\{\begin{array}{l}x_1-x_3=0\\x_2-3x_3=0\end{array}\right.$$and searching for solutions in $V$. You will get that$$\ker(f)=\left\{\begin{bmatrix}t\\3t\\t\end{bmatrix}\,\middle|\,t\in K\right\}.$$It follows now from the rank-nullity theorem that $\dim\operatorname{Im}(f)=1$. So, and since$$f\left(\begin{bmatrix}1\\1\\0\end{bmatrix}\right)=\begin{bmatrix}1\\1\end{bmatrix},$$you have$$\operatorname{Im}(f)=\left\{\begin{bmatrix}t\\t\end{bmatrix}\,\middle|\,t\in K\right\}.$$Finally, it is clear that$$\left\{\begin{bmatrix}1\\3\\1\end{bmatrix}\right\}\quad\text{and}\quad\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}$$are bases of $\ker(f)$ and of $\operatorname{Im}(f)$ respectively.
A: The equation defining the subspace $V$ tells us that $x_1=x_2-2x_3$, where $x_2$ and $x_3$ are free variables. (In other words, $V$ is the subspace that is the solution set of this linear equation.) So we can rewrite the definition of $V$ as
$$\begin{aligned}
V &= \left\{\begin{bmatrix} x_2-2x_3 \\ x_2 \\ x_3 \end{bmatrix} \colon x_2,x_3\in K\right\} = \\
&= \left\{x_2\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \colon x_2,x_3\in K\right\} = \\
&= \operatorname{Span}\left\{\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}\right\}\subseteq K.\end{aligned}$$
Then plugging in a generic vector from $V$ into the map $f$, and using $x_1=x_2-2x_3$, we get
$$f\left(\begin{bmatrix} x_2-2x_3 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix} (x_2-2x_3)-x_3 \\ x_2-3x_3 \end{bmatrix} = \begin{bmatrix} x_2-3x_3 \\ x_2-3x_3 \end{bmatrix} = (x_2-3x_3) \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$
This last line shows everything we wanted to know. We can see that the image is one-dimensional, the span of $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. And we can see that a vector in $V$ belongs to the kernel of $f$ iff $x_2-3x_3=0$, from which it's easy enough to find a description of this kernel.
