Given $$ A=\begin{pmatrix} -1 & -2 & 3 & -4 & -5 \\ 3 & 6 & -1 & 4 & 2 \\ -2 & -4 & 0 & -2 & 0 \\ -2 & -4 & 1 & -3 & 1 \\ \end{pmatrix} $$ and in solving for a basis for the null space of $A$, I found that: $$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ \end{pmatrix}=x_2 \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} + x_4\begin{pmatrix} -1 \\ 0 \\ 1 \\ 1 \\ 0 \\ \end{pmatrix}$$
Thus, a basis for the null space of A is $$B_A=\begin{Bmatrix} \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix},\begin{pmatrix} -1 \\ 0 \\ 1 \\ 1 \\ 0 \\ \end{pmatrix} \end{Bmatrix} $$
My question is, how would I find a new basis for the null space of $A$, whose vectors are NOT multiples? Are there any fast ways of doing this?
I am aware that you can use a change of basis matrix, but if a question asks you to find this new basis for null space and then find the change of basis matrix (not the other way around), how would this be done?
You probably want to find an orthonormal basis. Look into the Gram-Schmidt process.