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I'm going over the proof of the spectral theorem for compact symmetric operators in Hilbert space in Lax. Let $A$ be a compact symmetric operator on a Hilbert space to itself. Define the Rayleigh quotient to be

$$R_A(x) = \frac{(Ax, x)}{\|x\|^2}$$

Let $z$ be vector that maximizes the quadratic form $(Ax, x)$ over the unit ball. Let $w$ be arbitrary. Now the text claims that $R(z + tw)$ is differentiable, and since it achieves its maximum at $t=0$, it's $t$-derivative is zero, and we have

$$\frac{(Aw, z) + (Az, w)}{\|z\|^2} - (Az, z) \frac{(w, z) + (z, w)}{\|z\|^4} = 0$$

I don't see why the given function is differentiable, nor the computation of its derivative.

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    Consider finite dimensional case, for example2012-01-23
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    I don't think the argument is simpler in the finite-dimensional case. Instead, observe that the given function of $t$ can be written as the quotient of two real-valued functions of $t$, each of which is easily seen to be differentiable (just expand out the inner products, using bilinearity of $t$)2012-01-23
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    I am in a rush, perhaps I will write this up as an answer later, if time permits2012-01-23

1 Answers 1

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Since $t$ and $w$ are fixed, we can put $f(t):=R(z+tw)$ and as @Yemon Choi suggests, after expanding the inner products: $$f(t)=\frac{||w||^2t^2+2t\Re (Aw,z)+(Az,z)}{||w||^2t^2+2t\Re (z,w)+||z||^2}.$$ We have, using the classical rules of differentiation: $$f'(t)=\frac{2t||w||+2\Re (Aw,z)}{||w||^2t^2+2t\Re (z,w)+||z||^2}-\frac{||w||^2t^2+2t\Re (Aw,z)+(Az,z)}{(||w||^2t^2+2t\Re (z,w)+||z||^2)^2}(2t||w||^2+2\Re (Aw,z)),$$ so evaluating it at $t=0$ we get $$\frac{2\Re (Aw,z)}{||z||^2}-2\frac{(Az,z)}{(||z||^2)^2}\Re (Aw,z)=0,$$ which is what is wanted.