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Math Help - idempotent matrices

  1. #1
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    idempotent matrices

    idempotent matrices-math.jpg

    Having a huge amount of trouble with this question, been at it all afternoon. It is supposed to be relatively simple but im only in week 2 of this course and am baffled. Any help would be much appreciated.
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  2. #2
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  3. #3
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    I figured that xx' and x'x are the same thing (just x1^2+x2^2...xn^2) so xx'/x'x=1 ??

    therefore A=xx'/x'x is indepotent as A=1 and 1*1=1. subbing A=1 into B=In - A gives us B=In-1....am i on the right track?
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  4. #4
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    I'm afraid you're completely on the wrong track. xx' is a matrix, x'x is a number (it's the dot product). To convince yourself of this, look at the sizes of the vectors x (n x 1) against x' (1 x n), along with the definition of matrix multiplication and the size of the resulting multiplication.

    So, A is the matrix \frac{1}{x'x}\,xx'. What's in the denominator is a number: the dot product.

    I would ask yourself what the definitions of "idempotent" and "symmetric" are. What are those?
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  5. #5
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    Break the problem down into 2 parts. First, try to show that the matrix is symmetric (i.e. transpose(A) = A). Then, try to show that the matrix is idempotent (A^2 = A). Since you have shown both separately, it means both are true.

    Hint: You do not need to expand the vector x into its components to do this problem.
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