Proof that $X = \left\{ A \in \mathbb R^{6,5} : V \subset \ker(A) \right\}$ is linear subspace and find its $\dim$ Proof that if $\dim V = 3 $ then $X = \left\{ A \in \mathbb R^{6,5} : V  \subset  ker(A) \right\}$ is linear subspace and find its $\dim$
What I've done
I think that I have done proof.
$ \forall \vec{v} \in V. A\vec{v} = 0 $ and $B\vec{v}=0$
but from property of the matrix product it follows that: 
$$ A\vec{v} + B\vec{v} = (A+B)\vec{v} $$
Moreover if $A\vec{x}=0$ then $(\alpha A)\vec{x} =0$ and in to other side too. So ok.
My problem
But how in use of given informations find its dimension? My scribbles do not lead to anything.
 A: $\mathbb R^{6,5}$ is a linear space of dimension $6 \times 5=30$.
If $(v_1, v_2, v_3)$ (with $v_i \in \mathbb R^5$) is a basis of $V$. You have
$$X = \{A \in \mathbb R^{6,5} \ ; \ A.v_1=A.v_2=A.v_3=0\}.$$
It means that you're imposing $3 \times 6 = 18$ independent conditions in the dual space $\left(\mathbb R^{6,5}\right)^\vee$. Hence $\dim X = 30 -18=12$.
Another proof not using the dual space
Without loss of generality, you can suppose that $(v_1, \dots, v_5)$ is a basis of $\mathcal V = \mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^\prime, \dots, v_6^\prime)$ is a basis of $\mathcal V^\prime=\mathbb R^6$.
For $A =(a_{ij}) \in X$ expressed in those basis, you must have $$a_{11} =\dots =a_{61}=a_{12}=\dots=a_{62}=a_{13}=\dots=a_{63}=0$$ as $V \subseteq \ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, \dots, E_{64}, E_{15}, \dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.
A: Since $V\subset\ker(A)\forall A\in X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^\perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^\perp$.
$$A=\begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\\a_{21}v_1^T+a_{22}v_2^T\\\vdots\\a_{61}v_1^T+a_{62}v_2^T\end{bmatrix}=a_{11}\begin{bmatrix}v_1^T\\0\\\vdots\\0\end{bmatrix}+a_{12}\begin{bmatrix}v_2^T\\0\\\vdots\\0\end{bmatrix}+a_{21}\begin{bmatrix}0\\v_1^T\\\vdots\\0\end{bmatrix}+...+a_{62}\begin{bmatrix}0\\0\\\vdots\\v_2^T\end{bmatrix}$$
Thus, $X$ has $12$ basis vectors.
