Finding dimension or basis of linear subspace Let $\;F\;$ be some field and take a column vector $\;\vec x=\begin{pmatrix}x_1\\x_2\\\ldots\\x_m\end{pmatrix}\in F^m\;$ . We defined
$$W=\left\{\;A\in V=M_{n\times m}(F)\;:\;\;A\vec x=\vec 0 \;\right\}$$
It is asked: is $\;W\;$ a linear subspace of $\;V\;$ and if it is find its dimension.
This is what I did: if $\;A,B\in W\;,\;\;k\in F\;$ , then
$$\begin{align}&(A+B)\vec x=A\vec x+B\vec x=\vec 0+\vec 0=\vec 0\implies A+B\in W\\{}\\&
(kA)\vec x=k(A\vec x)=k\vec 0=\vec 0\implies kA\in V\end{align}$$
Since the zero matrix is in $\;W\;$ we get it is a linear subspace. 
They gave us hint for second part: that the answer to dimension of $\;W\;$ depends on $\;\vec x\;$ , which I can understand since if $\;\vec x=\vec0\;$ then $\;W=V\implies \dim W=nm\;$, but I can't see how to do in general case.
For example, if $\;\vec x=\begin{pmatrix}1\\0\\0\end{pmatrix}\in\Bbb R^3\;$ and $\;V=M_{n\times 3}(\Bbb R)\;$ , then
$$\begin{pmatrix}a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\\ldots&\ldots&\ldots\\a_{n1}&a_{n2}&a_{n3}\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\iff a_{11}=a_{21}=\ldots=a_{n1}=0$$
which gives us $\;\dim W=3n-n=2n\;$ , and if we take instead $\;\vec x=\begin{pmatrix}1\\1\\0\end{pmatrix}\;$ , then
$$a_{i1}=-a_{i2}\;\;\forall\;i=1,...,n\;$$ ...and again I get $\;\dim W=2n\;$ ! This confuses me, and any help with be very appreciated.

Here's one way to think about it: If $Ax = 0$, then each of the rows of $A$ is orthogonal to $x$.  The collection of vectors orthogonal to a nonzero vector is an $(m-1)$-dimensional subspace**.  Thus to specify a matrix $A$ with $Ax = 0$, we must choose $n$ rows each from this $(m-1)$-dimensional space, which implies that $\text{dim}(W) = n(m-1)$.  Note that this agrees with your examples, since you took $m = 3$.
**
We want to find the dimension of the subspace $U = \{a \in F^m : a_1 x_1 + \cdots + a_m x_m = 0\}$, where $x$ is a fixed nonzero vector.  Since $x \neq 0$ then $x_i \neq 0$ for some $i \in \{1, \ldots, m\}$.  Then
$$
a_i = \frac{1}{x_i} (-a_1 x_1 - a_2 x_2 - \cdots - a_{i-1} x_{i-1} - a_{i+1} x_{i+1} - \cdots - a_m x_m)
$$
so we may write
\begin{align*}
\begin{pmatrix}
a_1\\
a_2\\
a_3\\
\vdots\\
a_{i-1}\\
a_i\\
a_{i+1}\\
\vdots\\
a_m
\end{pmatrix} =
\begin{pmatrix}
a_1\\
a_2\\
a_3\\
\vdots\\
a_{i-1}\\
-\frac{1}{x_i} \sum_{\substack{k=1\\ k\neq i}}^m a_k x_k\\
a_{i+1}\\
\vdots\\
a_m
\end{pmatrix}=
a_1
\begin{pmatrix}
1\\
0\\
0\\
\vdots\\
0\\
-x_1/x_i\\
0\\
\vdots\\
0
\end{pmatrix}
+
a_2
\begin{pmatrix}
0\\
1\\
0\\
\vdots\\
0\\
-x_2/x_i\\
0\\
\vdots\\
0
\end{pmatrix}
+ \cdots +
a_m
\begin{pmatrix}
0\\
0\\
0\\
\vdots\\
0\\
-x_m/x_i\\
0\\
\vdots\\
1
\end{pmatrix} \, .
\end{align*}
Thus $U$ is spanned by the $m-1$ vectors $w_k$ with a $1$ in the $k^\text{th}$ spot and $-x_k/x_i$ in the $i^\text{th}$ spot, for $1 \leq k \leq m$, $k \neq i$.  One can easily show that these vectors are linearly independent, so $\text{dim}(U) = m-1$.
