What to do with an empty column in the basis of the null space?

The problem says to find a basis of the null space of A, A being the matrix:

$\begin{bmatrix}1 & 0 & 0\\1 & 0 & 1\end{bmatrix}$

So I need to solve the equation $Ax = 0$ to find the null space of A.

$\begin{bmatrix}1 & 0 & 0 \\1 & 0 & 1\end{bmatrix}$

reduces to

$\begin{bmatrix}1 & 0 & 0\\0 & 0 & 1\end{bmatrix}$

and I end with $x_1 = 0, x_3 = 0$

so this tells me the solution vector is

$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ = $\begin{bmatrix}0\\0\\0\end{bmatrix}$

Which is the zero vector. However the book the tells me the solution is:

$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ = $\begin{bmatrix}0\\1\\0\end{bmatrix}$

I do not know where they got the 1 from since $x_2$ was not a part of the system at all. If I had to take a guess I would say that because $x_2$ isn't included at all it can be anything since the result of the multiplication will always end up with $0$. Can anyone enlighten me on this? Am I close?

Thanks!

Solution 1:

This is a very common mistake.

You have no equation telling you about $x_2$. Lots of people assume this means $x_2$ must be zero.

On the contrary, since you have no information about $x_2$, it could be anything! So the solution is $$x_1=0\ ,\quad x_2=t\ ,\quad x_3=0$$ and the nullspace is $$\left\{t\pmatrix{0\cr1\cr0\cr}\ :\ t\in{\Bbb R}\right\}\ ,$$ which has a basis $$\left\{\pmatrix{0\cr1\cr0\cr}\right\}$$ as claimed.

Solution 2:

You shall end with $x_1=x_3=0, x_2=c$ for some constant $c$.