Answer
(a) Column space of $A$ is the set of all combinations $\alpha.\begin{pmatrix}1\\0\end{pmatrix}$ , $\alpha \in \mathbb{R}$ ,
null space of $A$ consists of all combinations $\beta.\begin{pmatrix}0\\1\end{pmatrix}$ , $\beta \in \mathbb{R}$ .
(b) Column space of $B$ is the set of all combinations $\alpha.\begin{pmatrix}0\\1\end{pmatrix}$+$\beta.\begin{pmatrix}3\\3\end{pmatrix}$ , $\alpha,\beta \in \mathbb{R}$ , null space of $B$ is zero .
(c) Column space of $C$ is zero , null space of $C$ is $\mathbb{R}$ .
Work Step by Step
Idea is checking if the vectors in the column space are dependent or not so we can eliminate dependent one to construct space from independent ones . For the null space , it's solving system of equations by multiplying matrix with unknown vector and trying to get zero .
