I'm working with the following kernel: $\ker\begin{pmatrix} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix}$, which I simplified to $\ker \begin{pmatrix} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $.
I want to find the basis of $\ker A$. I took "1" in the first row as the bound variable, and "-1" as the free variable, so that the first row has the equation: $x_2-x_3=0$, and assigned $x_3 = \alpha$. In the end I got only the base $\left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right)$, but I am supposed to find two basis. Where I did wrong?
Best Answer
As pointed out in the comments, $x_1$ is a free variable. So letting $x_1 = s$ and $x_3 = t$, an element of the kernel is of the form \begin{align*} \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix} &= \begin{pmatrix} s\\ t\\ t \end{pmatrix} = \begin{pmatrix} s\\ 0\\ 0 \end{pmatrix} + \begin{pmatrix} 0\\ t\\ t \end{pmatrix} = s \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + t \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix} \, . \end{align*} This shows that $(1,0,0)^T$ and $(0,1,1)^T$ span the kernel, and one can show that they are also linearly independent.