<s>[INST] Kernel, Basis and Column Space If a matrix $A$ has row reduced form
$A=$$\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$
then the kernel of the linear map $T : \mathbb{R}^4 → \mathbb{R}^3$ defined by $T(x) = Ax$ has basis $$(−3,1,0,0), (−3, 0, 1, 1).$$

How do I prove or disprove the claim regarding the kernel?

Any help is appreciated guys. Also, how can I show that columns 2 and 4 forms a basis for the column space?
 [/INST] The reduced matrix is $A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$ and if we want to find the generator of the Kernel we have to solve the system $A\underline x=\underline0$:
$$A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}\underline x=\underline 0\implies \begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \end{pmatrix}\begin{pmatrix}x\\y\\z\\w \end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies$$
$$\begin{cases}x+3y+3w=0\\z-w=0 \end{cases}\implies z=w=t\in\mathbb R,y=s\in\mathbb R,x=-3t-3s.$$
So we can state that $\operatorname{Ker}(F)=\Big\langle\begin{pmatrix}-3\\1\\0\\0 \end{pmatrix},\begin{pmatrix}-3\\0\\1\\1 \end{pmatrix} \Big\rangle$.
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