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Exercise 1.8.9

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$L = \frac{1}{2} x^{T} Sx + λ (x^{T} x - 1)$ , we have

\begin{array}{l} \frac{∂L}{\partial x_{i}} & = {(Sx)}_{i} + 2 λ {(x)}_{i} \end{array}

for $i = 1, 2, \dots, n$ .

So $\frac{∂L}{∂x} = Sx + 2 λx$ , let this equal to 0, we see that $Sx = -2 $λx$, so −2λ istheeigenvalueof S, and x isaneigenvectorof S .$

So when $x$ is the eigenvector of $S$ with eigenvalue $λ$ . $R (x) = \frac{x^{T} Sx}{x^{T} x} = x^{T} Sx = x^{T} = λ$ . This is maximum when $λ = λ_{1}$ .

niuers

2020-03-20 00:00

Exercise 1.8.9

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