Show that $W$ is a subspace and find its dimension. Let $x\in R^n$ be any nonzero vector. Let $W\subset R^{n\times n}$ consist of all matrices A such that $Ax=0$. Show that $W$ is a subspace and find its dimension.
It is trivial to verify that $W$ is a subspace. Since $A_1x=0$ and $A_2x=0$ implies that $(A_1+A_2)x=0$ and $kAx=0$ with $k\in R$.
I know that the dimension of vector space $F^{m\times n}$ of $m\times n$ matrices is $mn$. That is $E_{ij}$ which have a $1$ in the $i$-th row and $j$-th column and a zero everywhere else. When I tried to find the dimension of $W$ defined above, I cannot relate it with the rank of $A$.
This is my first blood on Math stack. Guys please conquer.
 A: Hint: For showing $W$ is a subspace, there are $3$ conditions to check.  Do you remember what they are?  They should be pretty straightforward to check so it would be better if you told us where you were stuck.
For the dimension: Look at the linear map $R^{n \times n} \to R^n$ given by $M \mapsto Mx$.  First off you should prove that this is linear.  Second, notice that the kernel of this map is exactly $W$.  Do you have a theorem about the dimensions of the kernel and image and domain of a linear map?
A: For proving $W$ is a subspace, you should also see that $0 \in W$ but it is trivial.
For calculating the dimension, let $T: \mathbb{R}^{n \times n} \to \mathbb{R}^n$ be defined by
$$ TA = Ax \qquad \text{ for } A \in \mathbb{R}^{n \times n} $$
Then by the dimension theorem,
$$ \dim(\mathbb{R}^{n \times n}) = \text{nullity}(T) + \text{rank}(T) $$
Note that $x \neq 0$. WLOG assume $x_1 \neq 0$. Then for any fix $k = 1, 2, \ldots, n$, we let
$$A_{k,1} = 1 \text{ and }A_{i,j} = 0 \text{ otherwise}$$
that is, 
\begin{equation*}
A = \begin{bmatrix}
0 & 0 & \ldots
& 0 \\
\vdots & \vdots & \ddots
& \vdots \\
1 & 0 & 0
& 0 \\
\vdots & \vdots & \ddots
& \vdots \\
0 & 0 & \ldots
& 0
\end{bmatrix}
\end{equation*}
Then we have 
$$Ax = (0, 0, \ldots, x_1, \ldots, 0)^T = x_1 e_k \neq 0$$
Note that $n$ different such matrices $A$'s are defined. Therefore we have $n$ independent vectors in $R(T)$ and 
$$ \text{rank}(T) = n $$
Therefore,
$$ n^2 = \dim(W) + n $$
$$\dim(W) = n(n-1) $$
