why not do this:

for i ≥ j define B = (b_{ij}) by b_{ij}= a_{ij}and for i < j, b_{ij}= a_{ji}.

clearly B is symmetric.

now let C = A - B. on or below the diagonal (i ≥ j), c_{ij}= a_{ij}- b_{ij}= 0.

thus C is upper triangular with all 0's on the diagonal.

example:

.

then:

.

it seems to me we can do this no matter what the eigenvalues of A are.

for 2: another way to write |x| is as the scalar:

√(x^{T}x). thus |Px| = √((Px)^{T}Px) = √(x^{T}(P^{T}P)x)

what is P^{T}P?