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Kate2010
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Homework Statement
T: V-> V, dimV = n, satisfies the condition that T2 = T
1. Show that if v [tex]\in[/tex] V \ {0} then v [tex]\in[/tex] kerT or Tv is an eigenvector for eigenvalue 1.
2. Show that T is diagonalisable.
Homework Equations
The Attempt at a Solution
I have shown in an earlier part of the question that the eigenvalues of T are 0 and 1.
1. Considering eigenvalue 1, we get T(v) = v, T(T(v)) = T(v), T(v) = T(v), this is true so T(v) is an eigenvector.
If v [tex]\in[/tex]kerT then we have T(v) = v, but then surely we get 0 = v which it isn't?
Also, I think I have worked back from the answer to show that these work, rather than showing that these are the only possible eigenvectors satisfying T(v) = v.
2. Suppose I have shown the 1st part correctly.
dim(kerT) +dim(Tv) = dim(kerT) + dim (imT) = dimV
So we have n linearly independent eigenvectors which means T is diagonal. I'm not so sure about this claim.