idempotent matrices

Aug 2010
2
0
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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.
 

Ackbeet

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Jun 2010
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What have you done so far?
 
Aug 2010
2
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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?
 

Ackbeet

MHF Hall of Honor
Jun 2010
6,318
2,433
CT, USA
I'm afraid you're completely on the wrong track. \(\displaystyle xx'\) is a matrix, \(\displaystyle x'x\) is a number (it's the dot product). To convince yourself of this, look at the sizes of the vectors \(\displaystyle x\) (n x 1) against \(\displaystyle x'\) (1 x n), along with the definition of matrix multiplication and the size of the resulting multiplication.

So, \(\displaystyle A\) is the matrix \(\displaystyle \frac{1}{x'x}\,xx'\). 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?
 
Aug 2010
3
0
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.