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## Homework Statement

If A, B are (complex) symmetric 3x3 matrices and det(xA-B) has three distinct solutions a,b and c, prove there exists P invertible such that

P

^{t}AP = I the identity matrix

P

^{t}BP = diag(a,b,c) the diagonal matrix with non-zero entries a,b and c

## Homework Equations

det(xA-B)=0 iff there exists v =/= 0 s.t. xAv = Bv

## The Attempt at a Solution

So we have three distinct vectors we'll call x,y and z such that

aAx = Bx

bAy = By

cAz = Bz

and from here we want to construct P. The first thing I notice is that Ax, Ay and Az are all non-zero, as otherwise Bx (or y or z) is too and one of a,b or c can be arbitrary (and hence they are not necessarily distinct). But then just because a, b and c aren't necessarily distinct doesn't mean that there aren't distinct a, b and c that do satisfy the condition. This way lies madness, so I decided to go another route:

If we naively try P = [x y z] then AP = [Ax Ay Az] and BP = [aAx bAy cAz] which looks like a pretty good start, but then P

^{t}doesn't do anything useful unless I'm missing something big.

Basically I'm just looking for what the right starting off point is