Does $ \vec x=\left(\begin{smallmatrix}1 \\ 0 \\ 1 \end{smallmatrix}\right)$ lie in the Image of $L$?

Question

I have the following linear transformation:

$$L: \mathbb{R^3} \to \mathbb{R^3}, \begin{pmatrix}a \\ b \\ c \end{pmatrix} \mapsto \begin{pmatrix}a-c \\ 0 \\ b \end{pmatrix}$$

and I have to check if $ \vec x=\begin{pmatrix}1 \\ 0 \\ 1 \end{pmatrix}$ lies in the Image of L. I know that the image is defined as $Im(L):={L(\vec v)\in\mathbb{R^3} | \vec v} \in\mathbb{R^3}$ but I can't figure out how to check if $\vec x$ lies in the image.

Answer 1

$L\begin{pmatrix}a \\ b \\ c \end{pmatrix}= \begin{pmatrix}1 \\ 0 \\ 1 \end{pmatrix}\implies a - c = 1, b = 1$. Take $(a,b,c) = (1,1,0)$, and the answer is yes ! Note that there are many choices because, say $c$ is any real number.