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Math Help - rank

  1. #1
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    rank

    let A be an nxn matrix st A^2 = A and rank(A) = n.

    then the dimension of A should be nxn= n^2 right?

    and i was wondering, if the rank(A) =n, then the det A= 1 ( since there will be n leading ones by row operations). then A is invertible.

    but should A be invertible, then shouldnt the dimA= n^2 be equal to the dim of the image which in this case is n? there seem to be a contradict
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  2. #2
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    I'm sorry, I don't understand what you mean by "dim(A)". A matrix is NOT a vector space and so does not have a dimension. It also does NOT follow that if A is invertible, then det(A)= 1. If A is invertible then dim(A) is not 0 but can be any non-zero number.

    If A is invertible, then its rank is equal to the dimension of its domain space and so is the rank of A^2, A^3, etc.
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  3. #3
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    hmm the question that i was given was " if A is an nxn matrix such that A^2 = A and rank(A) =n, prove that A is the identity matrix".

    and the answer to that:

    " the rank can be defined as the size of the largest non zero determinant of a matrix. if rank(A)=n, then det(A) is not zero, thus A is invertible. multiplying the given relation with A inverse on the left, gets A=I"

    which i dont really get..
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  4. #4
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    What part don't you get? If rank(A)= n, then A maps R^n onto all of R^n. That is, A is both one-to-one and "onto" and so is invertible. A^{-1}(A^2)= A^{-1}A so A= I. By trying to talk about "dim(A)", which makes no sense, you made the problem much harder!
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  5. #5
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    sorry.

    isint the rank= dim of the image?
    i dont get how the rank(A)= n means that A maps R^n onto R^n.
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