Re: Mathematical problem
Roshdy - I had that one a little bit off - at least "mirky" - I'll try again:
For any vector x, xT.x is the inner product, and is also a valid (possibly the ONLY) eigenvalue. So, let's assume that λ = xT.x
Now, the eigenvector associated with this λ is equal to e = 1/√λ . x
Equally, √λ . e = x
If √λ . e = x, then we must also have (λ is just a scaling factor) √λ . eT = xT
Each of the vectors, e and eT, are "off from x and xT" by the factor √λ which,
being the inner product, is already in a "squared nature".
(√λ . e).(√λ . eT) recovers x.xT