Consider a matrix space M as follows:
$$ M = \{ A \in R^{3x3};C(A) \subseteq span \{(1,2,3)^{T}\}\} $$
*( $C(A)$ meaning column space of matrix )
a) Find a base of M
b) Find a matrix $ A \in M $ such that $(1,1,-1)^T \notin Ker(A^T) $
Now with some help I was able to find a few matrices that are in M: $$ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 0 & 0 \\ 3 & 0 & 0 \\ \end{bmatrix} $$ as well as other matrices that have $(1,2,3)^T$ in other columns, but I have no idea how to continue further. The best I could think of is, that $$ base(M) = \{(1,2,3)^T\} $$ but I'm not sure if the base shouldn't be matrices as well. However, those are just speculations. All advice is strongly appreciated.
A) To find the base of M, we have to find all linearly independent matrices A such that $$ span(C(A)) = \{(1,2,3)^T\} $$ For that, we have to include only vectors $$ \vec{v} = (\alpha,2\alpha,3\alpha);\alpha\in \Bbb{R}$$ Because the dimension of vector space is $3$, base matrices should be also $3$. For example: $$ base(M) = ( \begin{bmatrix} 1 & 0 & 0 \\ 2 & 0 & 0 \\ 3 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 1& 0 \\ 0 & 2& 0 \\ 0 & 3& 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 3 \\ \end{bmatrix}$$
B) To find a matrix A $$ A\in M \text{ such that } (1,1,-1)^T \notin Ker(A^T) $$ We first take the general form of $A^T$ as a linear combination of base matrices $$ A^T = \begin{bmatrix} \alpha & 2\alpha & 3\alpha \\ \beta & 2\beta & 3\beta \\ \gamma & 2\gamma & 3\gamma \\ \end{bmatrix};\alpha,\beta,\gamma \in \Bbb{R}$$ But by multiplying this matrix by the mentioned vector we get $$ \begin{bmatrix} \alpha & 2\alpha & 3\alpha \\ \beta & 2\beta & 3\beta \\ \gamma & 2\gamma & 3\gamma \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0\alpha \\ 0\beta \\ 0\gamma \end{bmatrix} $$ which means that regardless of the chosen coefficients, the vector will always be in kernel of $A^T$, thus there's no such matrix A.