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.
Best Answer
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.