Trivial Eigenspace: Definition & Meaning

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What's a "trivial eigenspace"?
 
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Put it in context. Probably something to do with 0, and probably 0 as a vector.
 
"Let V be a vector space of dimension n over a field F, and let T:V->V be linear. Let R={linear transformations V->V that commute with T}.

Suppose T is diagonalisable, show that T is a commutative ring <=> all non-trivial eigenspaces of T are one-dimensional."

I just need to know what "non-trivial" means in this case. We defined the 0 vector to NOT be an eigenvector.
 
0 is not en eigenVECTOR, agreed, but as a subvector space 0 is an eigenSPACE, the trivial eigenspace.
 
How do I show that R is commutative <=> the minimal and characteristic polynomials of T are the same? (T might not be diagonalizable)
 
does that mean you've done the first question?

There is no harm in passing to the algebraic closure of your field if it helps you to think about it. Put everythin in jordan form, what can you say about the commuting matrices? just to give you some idea of how to use the difference between the minimal and characeristic poly.
 
Yes, I did the first question. I've got an idea about the second one that doesn't involve Jordan forms, but I'm still working on it.
 
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