Matrix space $ M = \{ A \in R^{3x3};C(A) \subseteq span \{(1,2,3)^{T}\}\} $ 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: 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.
