We have linear space of matrices of Real numbers of size $2 \times 2$; There exists a subspace of matrices $2 \times 2$, where $AX = XA$ ($X, A$ are matrices with size $2 \times 2$)
Is there any way I can imagine what does this subspace looks like in that linear space? Or any way how to visualize that subspace in that space?
Thanks
Well, this may not be a very satisfying answer, but ...
As usual, things get easier when you choose the right basis.
First, if $A$ is the zero matrix, then the subspace (which I'll call $S$) is all of $M_{22}$.
Second if $A$ has a single eigenvector for a nonzero eigenvalue, then there's an invertible matrix $Q$ with $$ Q^{-1} A Q = \pmatrix{k & 0 \\ 0 & 0} $$ Note that if $AX = XA$, then $$Q^{-1} A Q \cdot Q^{-1} X Q = Q^{-1} AX Q = Q^{-1} XA Q=Q^{-1} X Q \cdot Q^{-1} A Q$$.
In short, instead of looking for matrices that commute with $A$ in the standard basis, we can look for those that commute with $A' = Q^{-1} A Q$ in the $Q$ basis. Doing so yields exactly the diagonal matrices.
The preceding argument works because the Jordan normal form of $A$ is $A'$, and we can simply work in the Jordan basis. Well, there are three more cases to consider: (a) the matrix $A$ has two distinct eigenvalues, both nonzero, (b) the matrix $A$ has two identical nonzero eigenvalues, and is a multiple of the identity, (c) the matrix has one (repeated) nonzero eigenvalue, and its Jordan form has a "1" in the upper right corner and (d) the matrix has, as its Jordan form, $\pmatrix{0 & 1 \\ 0 & 0}$.
You can work through the analogous computations for each of these cases. Case b is easy: the set of commuting matrices is "all matrices". For case $d$, the answer is all matrices $\pmatrix{a & b \\ c & d}$ with $a = d$ and $c = 0$, i.e, the solution subspace is a product of the diagonal line in the $ad$-plane and the horizontal line in the $bc$-plane. Cases a and c are messier.