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Math Help - Sum of dimensions of eigenspaces?

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    Sum of dimensions of eigenspaces?

    How do I calculate the sum of dimensions of eigenspaces to evaluate whether or not a matrix is diagonalisable?

    E.g. dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) ?
    They're supposed to be written as vectors but I can't work out how to do that so I apologise for that. Thank you!
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    MHF Contributor Drexel28's Avatar
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by virussss123 View Post
    How do I calculate the sum of dimensions of eigenspaces to evaluate whether or not a matrix is diagonalisable?

    E.g. dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) ?
    They're supposed to be written as vectors but I can't work out how to do that so I apologise for that. Thank you!
    I'm confused. Can you restate the question.
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    MHF Contributor alexmahone's Avatar
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by virussss123 View Post
    How do I calculate the sum of dimensions of eigenspaces to evaluate whether or not a matrix is diagonalisable?

    E.g. dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) ?
    They're supposed to be written as vectors but I can't work out how to do that so I apologise for that. Thank you!
    dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) = 1 + 1 = 2
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    Re: Sum of dimensions of eigenspaces?

    I'm trying to find out if a matrix A is diagonalisable. In order to do this, I need to work out the sum of dimensions of eigenspaces.
    Does that make it clearer?
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by alexmahone View Post
    dim(span{(-2 0 1)}) + dim(span{(1 0 0)}) = 1 + 1 = 2
    Why does it end up as 1+1?
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    MHF Contributor alexmahone's Avatar
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by virussss123 View Post
    Why does it end up as 1+1?
    For any non-zero vector v, dim(span(v)) = 1 because a basis of span(v) is {v}, which contains only one vector.
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by alexmahone View Post
    For any vector v, dim(span(v)) = 1 because a basis of span(v) is {v}, which contains only one vector.
    That makes sense, thanks. So if there are 3 vectors, would it make the sum 3? i.e if there are 3 eigenvalues resulting in 3 different eigenspaces, would the sum of dimensions of eigenspaces be 3? Thank you
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    MHF Contributor alexmahone's Avatar
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by virussss123 View Post
    So if there are 3 vectors, would it make the sum 3?
    I guess so, as long as none of the vectors is the zero vector, in which case dim(span(v)) = 0.
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    Re: Sum of dimensions of eigenspaces?

    Quote Originally Posted by alexmahone View Post
    I guess so, as long as none of the vectors is the zero vector, in which case dim(span(v)) = 0.
    That answers everything. Thank you alexmahone!
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