Determine the null space of the matrix:$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$$
My try:
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}_{R_2\rightarrow R_2-2R_1\\R_3\rightarrow R_3-R_1}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}_{R_3\rightarrow 5R_3-2R_2}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 0 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{5}}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
From this I got $$x-y=0\implies x=y\\y=0$$
$$(x,y,z)^T=(y,0,z)^T=y(1,0,0)^T+z(0,0,1)^T$$
So, $(1,0,0)^T$ and $(0,0,1)^T$ is the null space. Is this correct?
Best Answer
Recall that by definition the nullspace is the subspace of all vectors $\vec x$ such that $A\vec x=\vec 0$ and in that case we have ony the trivial solution $(x_1,x_2)=(0,0)$ then $Null(A)=\{\vec 0\}$.
Notably to solve $Ax=0$ we can proceed by RREF to obtain
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}\to \begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
that is
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
that is $x_1=x_2=0$.