Linear Algebra Subspace Proof My problem is Let $A$ and $B$ be fixed $2\times 2$ matrices. Prove that the set $W=\{X: XAB=BAX\}$ is a subspace of $M_{2,2}$. 
The hints I have are 


*

*Start with two arbitrary elements of $W$, $X$, and $Y$. Write down the conditions they satisfy. 

*Show that the sum of $X$ and $Y$ still satisfy that condition.

*Show that any scalar multiple of $X$ still satisfied that condition. 


I am really confused on the first hint, and don't really know where to start on this problem. Any help would be very appreciated!!
 A: Take X, Y two elements of W. Then they satisfy:
XAB=BAX, YAB=BAY. 
So, for the sum we have (X+Y)AB=XAB+YAB=BAX+BAY=BA(X+Y). 
That means that the sum X+Y belongs to W.
Now for the scalar multiply, (λΧ)(ΑΒ)=λ(ΧΑΒ)=λ(ΒΑΧ)=ΒΑ(λΧ).
That means that λΧ belongs to W. 
A: The problem you're trying to solve is cut from the same cloth as many others, which all go something like this: "Here is a subset $W$ of a vector space $V$. Show that $W$ is in fact a subspace." To do this -- that is, to show that $W$ is a subspace -- there are only two things you have to check.


*

*Whenever one vector is in $W$, every scalar multiple of that vector is in $W$.

*Whenever two vectors are in $W$, their sum is in $W$.


As an example, consider the following three subsets of $M_{2,2}$:


*

*$W_1=\big\{X\in M_{2,2}:\text{at least one of the entries of $X$ is zero}\big\}$.

*$W_2=\big\{X\in M_{2,2}:\text{all of the entries of $X$ are positive}\big\}$.

*$W_2=\big\{X\in M_{2,2}:\text{the diagonal entries entries of $X$ are zero}\big\}$.


$W_1$ satisfies property 1 but not property 2 (why?). $W_2$ satisfies property 2 but not property 1 (why?). And $W_3$ satisfies both properties 1 and 2, meaning that only $W_3$ is a subspace.
