Computing the basis of a subspace of matrices with the given nullspace. Compute the dimension and find a basis for the subspace of matrices $H ⊂ M_{3×5} (\mathbb{R})$
such that $[0,2,−3,0,1]^T$ is in the nullspace of A ∈ H.
I have said that, if we want $[0,2,−3,0,1]^T$ to be in the nullspace of A, then we must be able to write,
$$c_1\vec{v_1}+...+c_k\vec{v_k} = [0,2,−3,0,1]^T $$
where $S = \{\vec{v_1},...,\vec{v_k}\}$ is a basis for the nullspace of A.
so we require
$$V\mathbb{\vec{c}} =  [0,2,−3,0,1]^T $$ to be consistent. where $V$ is a matrix whose columns are the elements of S.
How can I move on from here?
I feel as though this is wrong in the first place. Could someone give me push please.
 A: Consider a simpler case where you want a basis for the space $H$ of 2x3 matrices $A=(a_{ij})$ such that $2a_{i2}-a_{i3}=0$.  Then a general matrix in $H$ looks like
$$ \left( \begin{array}{ccc} a &b&2b \\ c &d & 2d \end{array}\right)$$
This equals 
$$ a \left( \begin{array}{ccc} 1 &0&0 \\ 0 &0 & 0 \end{array}\right) + b \left( \begin{array}{ccc} 0 &1&2 \\ 0 &0 & 0 \end{array}\right)+ c \left( \begin{array}{ccc} 0 &0&0 \\ 1 &0 & 0 \end{array}\right)+d\left( \begin{array}{ccc} 0 &0&0 \\ 0 &1 & 2 \end{array}\right)$$
and all of the four matrices above are in $H$.  Therefore they are a spanning set for $H$.  You can also see that they are linearly independent: if the expression above equals the zero matrix then $a=0$ (look at the top left corner), $b=0$ (the 1,2 entry), and so on.  Thus they are a basis of $H$, and it has dimension 4.
Your problem is similar but a bit harder, nevertheless you can do it a similar way.
A: $$\{\left( \begin{array}{ccc}
0 & -1 & 1 & 0 & 5\\
1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\end{array} \right),\left( \begin{array}{ccc}
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\end{array} \right),\left( \begin{array}{ccc}
0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 1\\
1 & 0 & 0 & 0 & 0\end{array} \right)
\left( \begin{array}{ccc}
0 & 0 & 0 & 0 & 0\\
0 & -1 & 1 & 0 & 5\\
0 & 0 & 0 & 1 & 0\end{array} \right)\left( \begin{array}{ccc}
1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 1\end{array} \right),\left( \begin{array}{ccc}
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & -1 & 1 &0 & 5\end{array} \right)\}$$
so $Dim(H) =6$
