How to find dimension of vector space In $\mathbb{R}^5$ there is given vector space $V$. Its dimension is 3. In $\mathbb{R}^{6,5}$ consider the subset $X = \{A \in \mathbb{R}^{6,5} : V \subset \ker A\}$. I have to show that $X$ is a vector space in $\mathbb{R}^{6,5}$ and find its dimension. To show that $X$ is vector space consider $x_1, x_2 \in X$ and $v \in V$. We know that $x_1 v = 0$ and $x_2 v = 0$ so $(\alpha x_1 + \beta x_2) v = \alpha (x_1 v) + \beta (x_2 v) = 0$ so $V \subset \ker (\alpha x_1 + \beta x_2)$. But I don't know how to find $X$'s dimension. Any ideas?
I understand $V \subseteq \mathbb{R}^5$ is a subspace, $\dim V = 3$ 

$X = \{A \in \mathbb{R}^{6 \times 5} : V \subseteq \ker A \}$

To show that $X$ is a vector space, it suffices to show it is a subspace of $\mathbb{R}^{6 \times 5}$.  


$0 \in X$, clearly because $V \subseteq \mathbb{R}^5 = \ker 0$
For $\alpha_i \in \mathbb{R}$ and $A_i \in X$, if $v \in V$, $A_i v = 0$, and so $ (\sum_i \alpha_i A_i) v = 0$. 


So it is a subspace.

Let $B = \{v_1, v_2, v_3 \}$ be a basis of $V$, and extend it to a basis $\{v_1, v_2, v_3, v_4, v_5 \}$ of $\mathbb{R}^5$.  As $Av_i = 0$ for $1 \leq i \leq 3$, you only have to say where goes $A v_i$ for $i=4$ and $5$.

So you have 5-3=2 degrees of freedom in the domain and 6 in the codomain, that gives $2 \cdot 6 = 12$.  I suspect the dimension is 12. Hope that helps.

I add this: you can think of $A$ as the matrix representation of some linear transformation $f: \mathbb{R}^5 \to \mathbb{R}^6$ with respect to the bases the extended version of $B$ above for $\mathbb{R}^5$ and and the standard basis for $\mathbb{R}^6$, so $f(v_i) = 0$ for $1 \leq i \leq 3$, and you can decide where goes $f(v_i)$ for $i=4,5$.
For the dimension of $X$, if $A\in X$, $A$ has to nullify $V$ and can do anything on the $2$-dimensional orthogonal complement of $V$. So the dimension of $X$ is the same as the dimension of the space of linear transformations from a $2$-dimensional space to a $6$-dimensional space, which is $12$.

EDIT: A commenter is unfamiliar with the direct bijection between $6\times 5$ matrices and linear transformations from $\mathbb{R}^5$ to $\mathbb{R}^6$;  we can reinvent like four or five wheels and say the same thing but longer:

Let $V$ have basis $\{v_1,v_2,v_3\}$. Use Gram Schmidt in $\mathbb{R}^5$ so that $\{v_4,v_5\}$ is a basis for $V^{\perp}$. Let $\{e_1,e_2,e_3,e_4,e_5,e_6\}$ be the standard basis for $\mathbb{R}^6$. 

Consider the matrices $A_{ij}$ defined as the matrix that takes $v_j$ to $e_i$ and all other $v_k$ to the zero vector in $\mathbb{R}^6$. Explicitly, $A_{ij}=C_{ij}P$, where $C_{ij}$ is a $6\times5$ matrix with all zeros except a $1$ in the $i,j$ position, and $P$ is the change of basis matrix from $\{v_1,v_2,v_3,v_4,v_5\}$ to the standard basis for $\mathbb{R}^5$. Explicitly, $P=\begin{bmatrix}|&amp;|&amp;|&amp;|&amp;|\\v_1&amp;v_2&amp;v_3&amp;v_4&amp;v_5\\|&amp;|&amp;|&amp;|&amp;|\end{bmatrix}^{-1}$. Check: $A_{ij}v_j=C_{ij}Pv_j=C_{ij}\epsilon_j=e_i$ where $\epsilon_j$ are the standard basis vectors in $\mathbb{R}^5$.

There are $30$ of these $A_{ij}$, which are matrices in $M_{6\times 5}$. They are linearly independent because if $\sum_{i,j} c_{ij}A_{ij}=[0]$, then multiplying against each of the $v_j$ one at a time gives $\sum_i c_{ij}e_i=[0]$, and since the $e_i$ are independent, the $c_{ij}$ must equal $0$.

But 18 of them are not in $X$, because they have one of $Av_1,Av_2,Av_3$ nonzero. Only the twelve matrices $A_{i4}$ and $A_{i5}$, which all have $Av_1=Av_2=Av_3=0$ are in $X$.

$M_{6\times5}$ is $30$-dimensional. We have established 12 linearly independent vectors that are in the space $X$. So $\dim V\geq12$. We have established 18 more linearly independent vectors that are not in $X$. So $\dim V\leq30-18-12$. So $\dim V=12$.