How to find a $3\times 3$ matrix with a certain null space and certain column space

Question

We were asked to

Find a $3\times 3$-matrix whose null space is the $x$-axis and whose column space is the $yz$-plane.

I was told that the answer is this, but I didn't understand why: $$\begin{bmatrix} 0&0&0\\0&1&0\\0&0&1 \end{bmatrix}.$$

Answer 1

Start out with a general matrix $$A:=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}.$$ We want its null space to be the $x$-axis. The $x$-axis is $$\{(x,y,z)\in\Bbb{R}^3\mid y=z=0\}=\{(x,0,0)\mid\ x\in\Bbb{R}\}=\operatorname{span}\{(1,0,0)\}.$$ So we want $Ae_1=(0,0,0)$, where $e_1=(1,0,0)$. Writing this out we find $$(0,0,0)=Ae_1=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=(a,d,g),$$ which gives us the first column of the matrix. Can you take the rest from here?