Linear Algebra Proof: Define matrix A_s := 1/2(A + A') where

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marylove

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Define a matrix A_s := 1/2(A + A') where A is a square matrix and A' its transpose. Show that:

. . .x'Ax = x'A_sx

I've tried all kinds of things...adding and subtracting the same things from the left side and trying to work towards the right hand side...direct substitution. I just can't get this. Can somebody shed some light? Thanks for any help.
 
What is x? Is x a nx1 column vector?
If so x<SUP>T</SUP>*A*x is a 1x1 matrix.
Also (x<SUP>T</SUP>*A<SUP>T</SUP>*x)<SUP>T</SUP>= x<SUP>T</SUP>*A*x.
 
I thought that I had given you the answer!
Because (x<SUP>T</SUP>*A<SUP>T</SUP>*x)<SUP>T</SUP>= x<SUP>T</SUP>*A*x; also (A+A<SUP>T</SUP>) is a symmetric matrix.
Can you prove that (x<SUP>T</SUP>*A<SUP>T</SUP>*x)= (x<SUP>T</SUP>*A*x)?
 
that is what I have been working on. I know that if I prove that then I am done. Is there any way to do it without summations? Because I don't understand how I can just switch the order of the sum when I am summing A_i,j in on and A_j,i entires in the other. Its not clicking to me.
 
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