Vtyjkyuk

Let

A=$beginbmatrix
2& 1& 0\
0& 2& 0\
0 & 0& 3
endbmatrix$

Which one is a subspace of vector space $M_3(mathbb R)$?

a) $Xin M_3:XA=AX$

b)$Xin M_3:X+A=A+X$

c)$Xin M_3:det(XA)=0$

My opinion that only c) is not a subspace because I can find two different matrices

$X_1$=$beginbmatrix
1& 1& 0\
0& 2& 1\
0 & 0& 0
endbmatrix$
$X_2$=$beginbmatrix
0& 1& 0\
0& 0& 1\
0 & 0& 1
endbmatrix$

that $det(X_1A)=0$ and $det(X_2A)=0$ but $X1+X2not in Xin M_3:det(XA)=0$
For b) it is subspace,easy because addition is commutative
For a) it is subspace, we have a subspace of matrices which look like this
$X$=$beginbmatrix
a& b& 0\
0& a& 0\
0 & 0& i
endbmatrix$, $a,b,iin mathbb R$ and for $forall x,yin Xin M_3:XA=AX$ and $forall alpha in mathbb R$, $x+yin Xin M_3:XA=AX$ and $alpha x in Xin M_3:XA=AX$. What you think?

One way to see it is by rewriting the defining relations into the following form
$$f(X)=O.$$
where $O$ can mean the zero element in any vector space ($Bbb R$, $M$ etc). Then decide if $f$ is linear.

In this way, the first set becomes the null set of $f(X)=AX-XA$, and the second $f(X)=A+X-X-A=O$, and the third $f(X)=det(AX)$.

As you can see, the first two $f$ are both linear and the third is not. So the first two are subspaces since they're kernels of linear maps. And then you just have to check the third one, which you already did.