Hey everybody,
This problem was on a homework that was already due... and I did it, but I did it in an incredibly convoluted way. So I was just wondering if anybody could find a better solution for it, since I'm sure one must exist:
Consider the vector space consisting of linear operators on V, some finite dimensional vector space. Given some member T of this vector space, we define Z(T) as the subspace consisting of all linear operators S that commute with T.
Show that the dimension of Z(T) is greater than or equal to Dim(V).
I had simplified the problem by considering Jordan canonical form, but then it just went downhill from there and got pretty messy. If anybody can find a more elegant way to do it, that'd be great!
EDIT: Whoops, forgot to mention that all vector spaces are over algebraically closed fields.
This problem was on a homework that was already due... and I did it, but I did it in an incredibly convoluted way. So I was just wondering if anybody could find a better solution for it, since I'm sure one must exist:
Consider the vector space consisting of linear operators on V, some finite dimensional vector space. Given some member T of this vector space, we define Z(T) as the subspace consisting of all linear operators S that commute with T.
Show that the dimension of Z(T) is greater than or equal to Dim(V).
I had simplified the problem by considering Jordan canonical form, but then it just went downhill from there and got pretty messy. If anybody can find a more elegant way to do it, that'd be great!
EDIT: Whoops, forgot to mention that all vector spaces are over algebraically closed fields.


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