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Diagonalisation of linear transformation watch

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    T: V-> V, dimV = n, satisfies the condition that T2 = T

    1. Show that if v \in V \ {0} then v \in kerT or Tv is an eigenvector for eigenvalue 1.

    2. Show that T is diagonalisable.


    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 \inkerT 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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    (Original post by Kate2010)
    T: V-> V, dimV = n, satisfies the condition that T2 = T

    1. Show that if v \in V \ {0} then v \in kerT or Tv is an eigenvector for eigenvalue 1.

    2. Show that T is diagonalisable.


    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 \inkerT 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.
    You've sorta done it, but in a very backwards way.
    Start from T=T^2. So T(v)=T(T(v)). Remember eigenvectors need to be non-zero.

    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.
    Mmm, you're pretty much there, maybe just be more explicit with your thinking. Anything in the kernel (bar zero) is an eigenvector (with eigenvalue zero). Anything in the image of T is also an eigenvector (as we've just shown to have eigenvalue one). So how many linearly idependent eigenvectors are there?
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    (Original post by Hathlan)
    You've sorta done it, but in a very backwards way.
    Start from T=T^2. So T(v)=T(T(v)). Remember eigenvectors need to be non-zero.



    Mmm, you're pretty much there, maybe just be more explicit with your thinking. Anything in the kernel (bar zero) is an eigenvector (with eigenvalue zero). Anything in the image of T is also an eigenvector (as we've just shown to have eigenvalue one). So how many linearly idependent eigenvectors are there?
    I'm still a bit confused about how to show for an eigenvalue of 1 that v could be in the kernel.

    Do we have n linearly independent eigenvectors as r(T) +n(T) = dimV = n?
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    ca=cb => c(a-b) = 0 => c = 0 or a=b
    Something similar may be useful here
 
 
 
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