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!
 A: 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.
A: You shall end with $x_1=x_3=0, x_2=c$ for some constant $c$.
A: The system of equations pins down what $x_1$ and $x_3$ equal, but don't say anything about $x_2$. That makes it a free variable that doesn't affect the others at all. In other words, $x_2$ can equal anything without making either equation false, so $\left[\begin{array}{c} 0 \\ x_2 \\ 0 \end{array}\right]$ solves the system no matter what value we take for $x_2$.
A: Your mistake is in concluding that $x_1=0,x_3=0$ means that $x_2=0$.
It does not follow. When you multiply the given matrix by a $3 \times 1$ column vector, none of your equations give you any constraint on $x_2$. That is, $x_2$ could be any number, and all the equations would be satisfied. That is where the basis vector is coming from.
