(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

T: V-> V, dimV = n, satisfies the condition that T^{2}= 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.

2. Relevant equations

3. 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.

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# Homework Help: Eigenvalues, linear transformations

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