What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$? 
If $\vec{x}$ is any vector in $\mathbb{R^n}$, what are the possible
  dimensions of the space $V$ of all $n\times n$ matrices $A$ such that
  $A\vec{x}=\vec{0}$?

I'm confused about a couple of things. First, isn't a dimension of the space $V$ the number of $n\times n$ matrices in a basis for $V$? So how do we speak of multiple dimensions for the same space? In terms of solving the problem, I think the answer is that $V$ is $n$ dimensional, since a basis could be $n$, $n\times n$ matrices such that the first one has the first row equal to $-x$, the second has the second row equal to $-x$, etc. (and all other rows equal to $\vec{0}$).
 A: The set of $n\times n$ matrices is itself a vector space (a different one than $\mathbb R^n$), because you can add matrices, and multiply them with a number, and they fulfil all the rules you need for a vector space. This vector space is often denoted $\mathbb R^{n\times n}$.
Now the space $V$ in the question is defined for a given vector $\vec x$ as the subspace of $\mathbb R^{n\times n}$ which fulfils $A\vec x=\vec 0$. This space obviously depends on the vector $\vec x$, that is, there is not just one such space, but a separate one for each vector $\vec x$ (for any pair of linear independent vectors, there exists a matrix that maps one of them to $\vec 0$, but not the other). This answers your question how you can have different dimensions: There are different spaces, one for each vector $\vec x$.
Actually what I've argued for above is strictly only that the sets $V$ are different; one would additionally have to show that each $V$ is indeed a subspace if $\mathbb R^{n\times n}$, which isn't that hard. However from the formulation of the question, I suspect you may take that as a given (probably because it was proven earlier in the course or book the given question origins from).
OK, so we have different spaces $V$, one for each vector $\vec x$, and therefore it makes sense for each of them to ask for the dimension. Now the dimensions cannot be arbitrary (for example, they obviously cannot be larger than the dimension of $\mathbb R^{n\times n}$. Indeed, the spaces $V$ for many vectors $\vec x$ will have the same dimension (but not all of them; think for example of the vector $\vec 0$).
The question now asks for all dimensions you can have for $V$ if $\vec x$ can be an arbitrary vector.
