I am not sure how to choose a basis for the kernel of a matrix. I've seen a video where someone chose for the kernel
$$\ker(A) = \ker\begin{pmatrix} -1 & 1 & 0 & 1 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
the basis
$$ \mathcal B =\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}\right\}$$
but I do not understand why those vectors were chosen.
Using the rank-nullity-theorem you can conclude there are two linear independent vector that span the $\ker(A)$ since ${\rm rank}(A)=2$ and $\dim(\ker(A))= n-{\rm rank}(A)=4-2=2$ but how do I choose those linear independent vectors?
Can anyone explain that to me?
Best Answer
Hint: You have already figured out that the kernel has dimension 2. The vectors $x$ in the kernel satisfy $-x_1+x_2+x_4=0$ and $4 x_3=0$.
You just need to find two vectors in the kernel that are linearly independent. They will form a basis of the kernel.