Determine if all $n \times n$ matrices such that $AB = BA$ for a fixed $n \times n$ matrix $B$ form a subspace of $M_{nn}$ I don't know how to start this one. Can I show that $A$ is closed under addition and scalar multiplication by saying for following:
 If $a, b \in \mathbb R$ and $A$ and $B$ are $n \times n$ matrices such that $AB = BA$ then $(aA + bB)B = aAB + bBB$
I'm not even sure this is legitimate what I have done so far. Some help starting would be appreciated.
Why in the duplicate question in the accepted answer does @Arturo say that you also need to show that there is at least one matrix such that $AB = BA$? I thought you only have to show closure under addition and scalar multiplication?
As has @Samuel noted below also.
 A: Your idea is great. First, you shouldn't call the two matrices that lie in the subspace $A$ and $B$ because the name $B$ is already taken. If you take two matrices $A_1,A_2$ in this set so that $A_1B = BA_1$ and $A_2B = BA_2$ and $a,b \in \mathbb{R}$ then
$$ (aA_1 + bA_2)B = aA_1B + bA_2B = aBA_1 + bBA_2 = B(aA_1 + bA_2) $$
which shows that this set is closed under addition and scalar multiplication (and it is clearly non-empty) so this is a vector subspace.
A: Hint:
I think you have the right idea, but you're messing up the notations.
What you have to prove is


*

*This set (let's denote it $S$ is not empty.

*If $A, A'$ are in $S$ and $a,a'\in\mathbf R$, then $aA+a'A'$ is in $S$.

A: Yes, $N = \{X\in M_{n,n} \mid XA = AX\}$ is a subspace of $M_{n,n}$. This follows from the fact that:


*

*The identity element of $M_{n,n}$, namely the zero matrix, is part of $N$ since $0A=A0$.

*Given any two matrices $X,Y\in N$ and two scalars $\alpha,\beta\in \mathbb{R}$, we have $\alpha X + \beta Y \in N$ because
$$A(\alpha X + \beta Y) = \alpha A X + \beta A Y = \alpha XA + \beta YA = (\alpha X + \beta Y)A.$$


Your reasoning was on the right track!
