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 $$A x = \lambda M x\tag{7.66}$$ 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 $$f(x) = A_{ij} x^i x^j\tag{7.67}$$ subject to the constraint that $$M_{ij} x^i x^j = 1.\tag{7.68}$$ 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 $E(x, \lambda)$ $$E(x, \lambda) = A_{ij} x^i x^j - \lambda (M_{ij} x^i x^j - 1).\tag{7.69}$$ Following earlier analysis, $$\frac{1} {2} \frac{\partial E(x)} {\partial x^i} = A_{ij} x^j - \lambda M_{ij} x^j.\tag{7.70}$$ Equating the partial derivatives to zero yields $$A_{ij} x^j = \lambda M_{ij} x^j\tag{7.71}$$ 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:
$$ A_{ij} x^j x^i = \lambda M_{ij} x^i x^j = \lambda \,.$$