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seen Mar 26 at 0:47

I'm an undergraduate student in Computer Science at the University of Waterloo.


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 suggested 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.