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Let $A\in M_n(R)$, where $M_n(R)$ is a set of all regular matrices with a rank $n$ and $A$ is a matrix from that set. Proof that the set of all matrices $X\in M_n(R)$, which commutates with $A$, is a vector space.

Well, the only matrix that commutates with $A$ is an identical matrix possibly multiplied with some scalar. How to prove it properly? Should I go from a definition, because it is pretty long, it has 7 or 8 parts. Thanks for helping me.

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    What makes you think the only matrices that commute with $A$ are those that are scalar multiples of the identity matrix? – 2017-02-06
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    right, all the matrices that are inverse to A also? – 2017-02-06
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    You don't need to know what are the matrices commuting to $A$. What you have to do is to check the axioms of a linear subspace. – 2017-02-06
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    Because the subspace is also a vector space, so as was written below (the answer by Chirantan Chowdhury): $(A+B)X = ... = X(B+A)$, it means $(A+B)$ is in that set and for $r$ from the real numbers (it is just a scalar) we have that $(rA)X = X(rA)$, so $(rA)$ in also in that set? – 2017-02-06

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Well not only the identity ,all the diagonal matrices with a fixed scalar $ \lambda$ on the diagonal shall commute with it.You don't need to exactly what are the matrices ,just see that if A,B are in the set , then$ (A + B) X = AX + BX = XB + XA = X(B+A) $ ,so $A +B$ is also in the set.Others also follow in the same manner.This is called the centralizer of $A$ .

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    Thank you very much for that help! – 2017-02-06