Vtyjkyuk

I'm working through Pavel Grinfeld's Introduction to Tensor Analysis and the Calculus of Moving Surfaces and I'm very stuck on exercise 118, which reads:

Show that the eigenvalues of the generalized equation (7.71) are given by the Rayleigh quotient
$$lambda = A_ij x^i x^j.$$

This references an equation from a preceding section of text:

In this section, we show that, much like $A x = b$, the eigenvalue problem
$$A x = lambda M xtag7.66$$
can be formulated as a variational problem. The matrix $M$ is assumed to be symmetric and positive define [sic]. The variational formulation is to find the extrema of
$$f(x) = A_ij x^i x^jtag7.67$$
subject to the constraint that
$$M_ij x^i x^j = 1.tag7.68$$
Geometrically, equation (7.68) states that the vector $x^i$ unit length [sic]. Use a Lagrange multiplier $lambda$ to incorporate the constraint on the augmented function $E(x, lambda)$
$$E(x, lambda) = A_ij x^i x^j - lambda (M_ij x^i x^j - 1).tag7.69$$
Following earlier analysis,
$$frac1 2 fracpartial E(x) partial x^i = A_ij x^j - lambda M_ij x^j.tag7.70$$
Equating the partial derivatives to zero yields
$$A_ij x^j = lambda M_ij x^jtag7.71$$
which is equivalent to the eigenvalue problem (7.66).

First I tried plugging the expression for $lambda$ (with renamed indices) into the RHS of equation (7.71) to try and obtain the LHS, but I didn't really get any further than plugging it in. Then I considered solving the variational problem for $lambda$ supposing we'd found an $x$ that works, but had no idea where to begin.

You can just multiply both sides of $(7.71)$ by $x^i$ and use the constraint:

$$ A_ij x^j x^i = lambda M_ij x^i x^j = lambda ,.$$