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I'm trying to prove that if there is $Z \in M_n(\mathbb{C})$ such that $[X, Z] = 0$ for all $X \in M_n(\mathbb{C})$, then $Z=cI$ for some complex number $c \in \mathbb{C}$.

I first noted that if $Z$ fulfills this requirement, then $Z + aI$ also does. This seems like the start of my proof. Now I'm not sure where to go from here. I know that I need to show that if $Z$ isn't a multiple of $I$, then it can't commute with every matrix. I'm not sure how/if I should leverage the first stated fact to do this. Can I get a hint?

sjgandhi2312
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    Your problem statement seems garbled in that $X$ should be a matrix, not a complex number (where you wrote $X\in \mathbb C$). In any case if I understand the problem you meant to ask, this has been asked before. See matrices that commute with all matrices – hardmath Aug 04 '21 at 14:59
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    Hint: Let $A_{ij}$ be a matrix with $1$ in entry $(i,j)$ and $0$ elsewhere – Matthew H. Aug 04 '21 at 14:59
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    Since this is tagged Representation Theory I'll point out that with a little cleverness you can do this directly via Schur's Lemma (i.e. your statement implies $Z$ commutes with a dim n irreducible representation of a finite group) – user8675309 Aug 04 '21 at 18:05
  • @user8675309 How can I use the fact that Z commutes to show that it makes an isomorphism from $M_n(\mathbb{C})$ to itself? Do we construct some new mapping that inputs an element from $M_n(\mathbb{C})$ to the commutator with $Z$ and then show that the result always being zero means we have some sort of irreducible representation? – sjgandhi2312 Aug 05 '21 at 14:13
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    'How can I use the fact that Z commutes to show that it makes an isomorphism from $_n$(ℂ) to itself?" You can't. This is not a group and $Z$ need not be invertible. E.g. $Z=\mathbf 0$ works and a priori you don't know there isn't some satisfying $Z$ with $0\lt \text{rank}\big(Z\big) \lt n$. The point is, with scalars over $\mathbb C$ there is always an irreducible $n$ dimensional (matrix) representation of a finite group. Why is this true? (Hint: consider $S_d$ the permutation group for operating on a set with $d$ elements). $Z$ commutes with said rep so you can apply Schur's Lemma. – user8675309 Aug 05 '21 at 17:27

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