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I have a question regarding one proof of the the following statement:

If $A$ is a square matrix such that $AX=XA$ for any square matrix $X$, then $A=cI$ for some $c$. ($I$ is the identity matrix).

I managed to prove this for $n \times n$-matrices when using the hint that I should let $X=E_{ij}$. However, I don´t really understand why that special case helps me prove it for all $n \times n$-matrices $X$.

By letting $X=E_{ij}$, we see that for $AE_{ij}=E_{ij}A$ to be true, we need $a_{ii}=a_{jj}$ and all the other entries at row $i$ and column $j$ to be zero. Since we may choose $i$ and $j$ arbitrary, this proves that $A=cI$.

But how can we now be sure that this holds for any matrix $X$? I understand Why it is a convenient approach to let $X=E_{ij}$ but don´t how it covers all cases? It feels like I´m missing something trivial since associating with $E_{ij}$ surley wasn´t obvious to me, even though it made the problem became "easy" to solve. That is not fine by me though, I really want to understand this.

alot
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  • possible duplicate of The Center of $\operatorname{GL}(n,k)$, resp. http://math.stackexchange.com/questions/27808/a-linear-operator-commuting-with-all-such-operators-is-a-scalar-multiple-of-the. – Dietrich Burde Feb 11 '15 at 14:40
  • I don't realy understand what is the problem. Yow already show that if $AX=XA$ for any $X$ then $A=cI$. The other direction is obvious. – Ofir Schnabel Feb 11 '15 at 14:40
  • You can note that the $E-{ij}$ form a basis. – Emilio Novati Feb 11 '15 at 14:52
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    You are not proving something about "all [matrices] $X$", but rather proving something about $A$. You have a hypothesis to work with that says $A$ commutes with all (square) $X$, and this is really equivalent to knowing $A$ commutes with all the $E_{ij}$, though you don't need to prove that equivalence. Just use the hint. – hardmath Feb 11 '15 at 15:19

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Keep in mind what you are proving.

If $A$ is a matrix such that $AX=XA$ for all matrices $X$, then $AX=XA$ for $X=E_{ij}$ (and some other matrices too).

You have shown that if $AX=XA$ for all $X=E_{ij}$ then $A = cI$ where $c=a_{11}=a_{22}=\cdots$.

So you have that if $A$ commutes with everything, then $A$ is a scalar multiple of the identity.

Now if you want to prove the converse, then you need to worry about all matrices.

But this is easy: Suppose $A=cI$. Then $AX=cIX=cX=Xc=XcI=XA$ so $A$ commutes with every matrix.

To sum up: You don't need to worry about limiting yourself to a certain subcollection of matrices in that part of your proof because you are (in this part) trying to determine what your matrix looks like.

To prove: "Commutes iff Scalar Multiple of Identity"...

Suppose Commutes with all. Therefore, commutes with some (specifically the $E_{ij}$'s). Thus is a Scalar Multiple of the Identity.

Then suppose it is a Scalar Multiple of the Identity. Then show it commutes with all matrices.

Bill Cook
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