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 Oct 13 awarded Popular Question Sep 24 awarded Autobiographer Mar 1 awarded Nice Question Mar 19 revised A linear operator commuting with all such operators is a scalar multiple of the identity. Minor wording improve. Mar 18 awarded Supporter Mar 18 awarded Scholar Mar 18 awarded Editor Mar 18 accepted A linear operator commuting with all such operators is a scalar multiple of the identity. Mar 18 revised A linear operator commuting with all such operators is a scalar multiple of the identity. Expanded some minor details. Mar 18 suggested approved edit on A linear operator commuting with all such operators is a scalar multiple of the identity. Mar 18 comment A linear operator commuting with all such operators is a scalar multiple of the identity. Thanks! Uniqueness can be shown by taking a basis $(v_{1},...,v_{n})$ and considering that $$Tv = T(a_{1} v_{1}+...+a_{n}v_{n}) = a_{1}b_{1}v_{1}+...a_{n}b_{n}v_{n}$$ but also $$Tv = kv = k(a_{1} v_{1}+...+a_{n}v_{n}) = a_{1}kv_{1}+...+a_{n}kv_{n}$$ Since Tv can be obtained in a unique way as a linear combination of said basis, then it follows that all the $b_i$ are equal to $k$. Hence, $Tv_{i} = kv_{i}$ and so $Tv = kv$, $\forall v$. Mar 18 comment A linear operator commuting with all such operators is a scalar multiple of the identity. It could very well be that I'm misunderstanding something, but doesn't the fact that T is a scalar multiple of I imply that Tv = av, where a is "fixed" and not dependent on v? Mar 18 awarded Student Mar 18 asked A linear operator commuting with all such operators is a scalar multiple of the identity.