Find a basis for the null space of a linear transformation T when we are given the spanning set of the range of T

Question

Let $T:\mathbb{R}^{5}\rightarrow\mathbb{R}^{4}$ be a linear transformation of the form $Tx=Ax$ where $A$ is a matrix of appropriate size with entries in the field of real numbers. Suppose that the range of T is spanned by the vectors $\alpha_{1}=\left(1,1,2,4\right),\alpha_{2}=\left(2,-1,-5,2\right),\alpha_{3}=\left(1,-1,-4,0\right),\alpha_{4}=\left(2,1,1,6\right)$ . Find a basis for the null space of $T$

Answer 1

The given vectors span the image of $$ P = \begin{bmatrix} 1 & 2 & 1 & 2 & 0 \\ 1 & -1 & -1 & 1 & 0 \\ 2 & -5 & -4 & 1 & 0 \\ 4 & 2 & 0 & 6 & 0 \end{bmatrix} $$ and the image of $$ Q = \begin{bmatrix} 0 & 1 & 2 & 1 & 2 \\ 0 & 1 & -1 & -1 & 1 \\ 0 & 2 & -5 & -4 & 1 \\ 0 & 4 & 2 & 0 & 6 \end{bmatrix} $$

while a basis for the kernel of $P$ is $$ \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ and a basis for the kernel of $Q$ is $$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$ showing that the basis cannot be determined uniquely from the information given.

(This is merely an amplification of G. Sassatelli's comment.)