Finding the kernel and image of a linear transformation.

$$L : \Bbb{R}^3 \to \Bbb{R}^2 : \begin{pmatrix} \alpha \\ \beta \\ \gamma\end{pmatrix} \mapsto\begin{pmatrix}\alpha + 2\beta + 3\gamma \\ 4\alpha + 5\beta + 6\gamma\end{pmatrix}$$

Show that this map is a linear transformation of $$\mathbb{R}$$ vector spaces. Then determine the core (kernel) and the image of $$L$$.

I have determined that the image is a linear transformation, but I'm stuck at the second part of the question. My question is then how do I determine the kernel and image? Perhaps you could give a more general approach since this is not the only task I have problems with.

• You compute the kernel of a linear map with matrix $A$ by solving $Ax=0$ via Gauss Elimination. If you find the conditions for $Ax=y$ to have a solution, again by Gauss Elimination, you'll get a set of linear equations on the coords of $y$: solve them via Gauss Elimination and you'll have the image. Commented Jul 14, 2020 at 15:29
• Take the canonical vectors $\{e_1,e_2,e_3\}$ which are a basis for $\mathbb{R^3}$ and find $L(e_1),L(e_2),L(e_3)$. Now you can consider the matrix $A=(L(e_1)|L(e_2)|L(e_3))$ and reduce it with Gauss: the columns which correspond to the non-zero pivot are a basis for $Im(L)$. Then, if you want to find the dimension of the kernel you can easily calculate it using $\bf{Grassman}$ formula and a basis of the Kernel can be found computing $Ax=0$, where $A$ is the representative matrix of your linear application. I tried to be general as you asked. Commented Jul 14, 2020 at 17:17
The transformation is the same as matrix multiplication by the matrix $$\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}$$, and is thus linear.
The image is the span of the columns. Or $$\rm{span}\{(1,4)^t,(2,5)^t\}=\Bbb R^2$$.
The $$1$$-dimensional kernel can be found by row reducing the matrix.