Find basis to $\begin{bmatrix}1&-4&3&-1\cr2&-8&6&-2\end{bmatrix}$ I want to find the basis to
$$x_1-4x_2+3x_3-x_4=0$$
$$2x_1-8x_2+6x_3-2x_4=0$$
so I set up the matrix:
$\begin{bmatrix}1&-4&3&-1\cr2&-8&6&-2\end{bmatrix}$
to get $\begin{bmatrix}1&-4&3&-1\cr0&0&0&0\end{bmatrix}$
Then I would get $\begin{bmatrix}x_1\cr x_2\cr x_3\cr x_4\end{bmatrix}=\begin{bmatrix}4b-3c+d \cr b\cr c\cr d\end{bmatrix}$. The issue is that I don't know how to then find the basis afterwards. Somone said I have to get $=b\begin{bmatrix}4\cr 1\cr 0\cr 0\end{bmatrix}+c\begin{bmatrix}-3\cr 0\cr 1\cr 0\end{bmatrix}+d\begin{bmatrix}1\cr 0\cr 0\cr 1\end{bmatrix}$ and each of the matrices are the vectors of the basis set. But I wasn't sure howe we get the last step?
 A: You find a basis for a vector space, not for a linear equation.
According to what you did, you are looking for a basis for the kernel of the coefficient matrix, or equivalently, a basis for the solution space of the homogeneous linear equation.
You have already found an answer:
$\left\{\begin{bmatrix}4\cr 1\cr 0\cr 0\end{bmatrix},
\begin{bmatrix}-3\cr 0\cr 1\cr 0\end{bmatrix},
\begin{bmatrix}1\cr 0\cr 0\cr 1\end{bmatrix}\right\}$
A: Hint : First of all, let's recall the definition of basis of a subspace : A set of linearly independent vectors which span the subspace,is called a basis of the subspace. 
You have found the basis for Null space of $A$. To prove that the set of vectors you have found is indeed a basis, you must only show that :
$\{(x_1,x_2,x_3,x_4)^T\in \mathbb R^4: (x_1,x_2,x_3,x_4)^T $satisfies the given system of linear equations $\} = \operatorname{span} (p, q, r) $, where $p, q, r$ are the vectors which you have found. 
In order to show that, let $y\in $LHS and show that $y\in $ RHS so that $LHS\subseteq RHS$ and similarly show the converse. This will prove the result.
