Find $L(\vec e_3)$ such that $\vec e_3 \in Kern(L)$.

Question

I have the following linear transformation:

$$L: \mathbb{R^3} \to \mathbb{R^4} $$ $$L(\begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix}) = \begin{pmatrix}1 \\ 0 \\ 2 \\0 \end{pmatrix}$$ and $$L(\begin{pmatrix}0 \\ 2 \\ 0 \end{pmatrix}) = \begin{pmatrix}0 \\ -1 \\ 0 \\-3 \end{pmatrix}$$

and I have to find $L(\vec e_3)$ such that $\vec e_3 \in Kern(L)$. I feel that I have to somehow change the original transformation into an algebraic one in order to work out $L(\vec e_3)$ but I'm not sure.

Answer 1

In order for $v$ to be in the kernel of a linear transformation, it must evaluate to the zero vector in the image of the transformation. Thus, if $e_3\in\ker(L)$ we must have $L(e_3)=0$.