I'm really confused here. I'm having trouble seeing why is $ x_1 $ a free variable instead of a zero-valued variable, while finding the null space of the following matrix. I shall post my progress:
This is the system in the form Ax=0, with A already in RREF :
$\begin{bmatrix}0 & 1 & -3\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$ $\begin{bmatrix}x_1 \\x_2\\ x_3\end{bmatrix}$ = $\begin{bmatrix}0 \\0\\ 0\end{bmatrix}$ which implies that $ 0x_1 +x_2-3x_3=0 $
My mistake is taking $x_1$ as zero, then with this assumption a basis for the Null Space would be: $\begin{bmatrix}0 \\3x_3\\ x_3\end{bmatrix}$ = $ \begin{bmatrix}0 \\3\\ 1\end{bmatrix}$
But this is wrong because $x_1$ is a free variable apparently, so the right way for finding a basis for the said Null Space would be: $\begin{bmatrix}0 \\3x_3\\ x_3\end{bmatrix}$ + $\begin{bmatrix}x_1 \\0\\ 0\end{bmatrix}$ , which leads to the correct basis vectors $ \begin{bmatrix}0 \\3\\ 1\end{bmatrix}$ ; $ \begin{bmatrix}1 \\0\\ 0\end{bmatrix}$
I don't understand why it should be this way. It is driving me insane and I have not found an answer on the internet nor in my book so far (this appeared on a test and I couldn't assist to the revision with my professor).
I've designed other matrices with the same form and I realized that this also holds true, for example: $\begin{bmatrix}1 & 1 & 0 \\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$ has a basis of two vectors $(-1,1,0)$ and $(0,0,1)$ so the same pattern occurs. Thank you in advance
Your equation $0x_1+x_2−3x_3=0$ is correct. Note $x_1=0,x_2=3,x_3=1$ is a solution: $$0\cdot 0+3−3\cdot 1=0$$ But $x_1=1,x_2=3,x_3=1$ is also a solution $$0\cdot 1+3−3\cdot 1=0$$ And $x_1=2,x_2=3,x_3=1$ is also a solution $$0\cdot 2+3−3\cdot 1=0$$ In fact any $x_1$ , $x_2=3,x_3=1$ is a solution $$0\cdot x_1+3−3\cdot 1=0$$ So why would you assume $x_1=0$? The nullspace is just the set of all $[x_1,x_2,x_3]^T$ for which $0x_1+x_2−3x_3=0$, a condition which does not constrain the value of $x_1$ at all.