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

  • Thread starter Thread starter UD1
  • Start date Start date
  • Tags Tags
    Kernel
UD1
Messages
19
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top