Finding the Dim of a matrix $A $ such that $AC = 0$ 
Hi,
So the answer that is given to this problem is $n(n-1)$, which doesn't make a lot of sense to me. 
Take for example a $3\times3$ matrix: 
$$ 
A=\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{33} \\
a_{31} & a_{32} & a_{33} \\ 
\end{bmatrix}
$$ 
$$ 
A C=\begin{bmatrix}
c_1\times a_{11} + c_2\times a_{12} + c_3\times a_{13} \\
c_1\times a_{21} + c_2\times a_{22} + c_3\times a_{33} \\
c_1\times a_{31} + c_2\times a_{32} + c_3\times a_{33} \\ 
\end{bmatrix}
$$ 
If $C_1,C_2 $or $C_3 = 0$ then dim($A$) can be  $0,1,2,3 $
If $C_1$ and $C_2 = 0$ them Dim($A$) can be $0,1,2,3,4,5,6$
etc.
If $C_1= C_2 = C_3 = 0$ Dim($A$) can be $9$ 
Where am I going wrong? 
 A: We can think at an $ n \times n $ matrix $A$ as an element of a vector space $V$ of dimension $n^2$. We can figure this thinking at $A$ as a set of $n$ rows $\mathbf{a_i}$ each of them have $n$ independent elements $a_{ij}$, so that matix can be expressed as a linar combination that have as coefficients these $n^2$ numbers and as basis the $n\times n$ matrices that have only one entry $=1$ and all other entries null.
Now, your question asks what is the dimension of the subspace of $V$ of all matrices $A$ such that $A\mathbf{c}=0$ for any $c$.
This means that for each row we can chose $n-1$ entries, but the last entry $ a_{in}$ is determined by the condition :
$$
c_na_{in}=-\sum_{j=1}^{n-1}a_{ij}c_j
$$ 
so for each of the $n$ rows we can chose only $(n-1)$ indpendent values, and the dimension of the subspace is $n (n-1)$.

In your example you have to consider $c_1,c_2,c_3$ all $\ne 0$ and you can see that:
$$
c_1\times a_{11} + c_2\times a_{12} + c_3\times a_{13}=0
$$
gives $a_{13}$ as a function of the other elements...and so one for the other rows.....
