Is the set of all $n\times n$ matrices, such that for a fixed matrix B  AB=BA, a subspace of the vector space of all $n\times n$ matrices? 
Is the set of all $n\times n$ matrices, such that for any fixed matrix B AB=BA, a subspace of the vector space of all $n\times n$ matrices?

Alright, I understand the question and I know what I have to do, basically. I need to show that additive closure and multiplicative closure are satisfied. The problem is, I can't figure out how to do this generally. I tried playing around with $2\times 2$ matrices but that seemed like a dead end. Obviously two such matrices are the 0 matrix and the identity matrix, and those form a subspace, but that doesn't really tell me about all the matrices. Any ideas for how I should be tackling this? I feel like I'm not thinking generally enough. 
 A: Careful: when you say "multiplicative closure", you have to be clear, since when dealing with $n\times n$ matrices there is a "multiplication" that has nothing to do with the vector space structure (the multiplication of matrices). It is clearer if you refer to it as the "scalar multiplication". 
So, fix the matrix $B$. You need to show that:


*

*There is at least one matrix $A$ such that $AB=BA$;

*If $A_1$ and $A_2$ are two matrices such that each of them commutes with $B$, then $A_1+A_2$ also commutes with $B$ (closure of your set under vector addition).

*If $A$ is a matrix that commutes with $B$, and $k$ is any scalar, then $kA$ also commutes with $B$ (closure of your set under scalar multiplication).


So, with that in mind:


*

*Is there a matrix that necessarily commutes with $B$? (If this thing is really going to be a subspace, it better have what "vector" [i.e., matrix] in it for sure? Try that matrix).

*Suppose $A_1$ and $A_2$ both commute with $B$. That is, $A_1B=BA_1$ and $A_2B=BA_2$. You want to show that $(A_1+A_2)$ also commutes with $B$. That is, you want to show that
$$(A_1+A_2)B = B(A_1+A_2).$$
Of course, you'll want to use the fact that each of $A_1$ and $A_2$ commutes with $B$, and perhaps some properties you know about matrix multiplication. Is there some property of matrix multiplication that would let you relate $(A_1+A_2)B$ with the products you do know something about, namely $A_1B$ and $A_2B$?

*Suppose $A$ commutes with $B$, and $AB=BA$. Let $k$ be a scalar. You want to show that $kA$ also commutes with $B$: that is, you need to prove that
$$(kA)B = B(kA).$$
Again, is there some property of matrix multiplication that you know and that might help here?
And if you establish these three, you're done: the set in question is a subspace!
A: Assume you have 2 such matrices, $A_1,A_2$ then 
$$(A_1+A_2)B = B(A_1+A_2)$$
must hold, so you need to prove this, using the fact that $A_iB=BA_i$ for $i=1,2.$
You then need to show that if $AB=BA$ the it also holds that $(\lambda A)B=B(\lambda A)$.
Tip: Use distribution law for matrices...
A: Here is a different kind of answer that is just a lot harder for no reason, but might help you transition.
Suppose we had a specific B, such as:
$$B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$
We want to find all A so that AB = BA.  Well, we don't know the entries of A, so we replace the unknown numbers with... well unkowns, also known as variables.
$$A = \begin{bmatrix} x & y \\ z & t \end{bmatrix}$$
Ok, now we write down what we know about the variables from AB = BA:
$$
AB =
\begin{bmatrix} x & y \\ z & t \end{bmatrix}
\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = 
\begin{bmatrix} 2x+4y & 3x+5y \\ 2z+4t & 3z+5t \end{bmatrix}
$$
$$
BA =
\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}
\begin{bmatrix} x & y \\ z & t \end{bmatrix} = 
\begin{bmatrix} 2x+3z & 2y+3t \\ 4x+5z & 4y+5t \end{bmatrix}
$$
For two matrices to be equal, we need to have the entries equal:
$$\left\{\begin{align}
2x + 4y &= 2x + 3z \\ 2z+4t &= 4x+5z \\ 4x+5y &= 2y + 3t \\ 3z+5t &= 4y+5t \end{align}\right.$$
What do you know?  We have a system of linear equations.  We can solve them as usual: find a particular solution (oooo let me let me!): $(x=0,y=0,z=0,t=0)$.  Now we know all other solutions are found by adding a "homogeneous solution", which form a subspace.
If we want, we can be even more linear-algebra-y.  let's move all the variables to one side:
$$\left\{\begin{align}
0x + 4y - 3z + 0t &= 0 \\ -4x+0y-3z+4t &= 0 \\ 4x+3y +0z -3t &= 0 \\ 0x-4y+3z+0t &= 0 \end{align}\right.$$
We can write it as a matrix equation:
$$\left[\begin{array}{rrrr}
0 & 4 & -3 & 0 \\ -4 & 0 & -3 & 4 \\ 4 & 3 & 0 & -3 \\ 0 & -4 & 3 & 0 \end{array}\right] \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
So we are just asking for the null space of the matrix!  Clearly that is a subspace.  We could even do Gaussian elimination to find which one.  I won't.
