Kernel of GL(n,F) acting on F^n

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Homework Statement



Suppose GL(n,F) acts on F^n in the usual way. Consider the induced action on the set of all k-dimensional subspaces of F^n. What's the kernel of this action? Is it faithful

The Attempt at a Solution



Well, I anticipate that the kernel of this action consists of scalar matrices, that is scalar multiples of the identity matrix. The question is how to prove that if g in GL(n,F) is a not a scalar matrix then we can always construct a k-dimensional subspace not fixed by g.
 

Answers and Replies

  • #2
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If A is not a multiple of the identity then there is a f such that Af is not proportional to f. Now, assuming k<n, you can construct a k-dimensional subspace such that f is in this subspace and Af is in a complementary subspace. For instance you extend the pair f,Af to a basis and construct your subspaces out of the basis vectors.
 

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