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Linear Algebra and Learning from Data

Exercise 1.7.28

Answers

1.
$λ_{1} I - S = λ_{1} I - QΛ Q^{T} = Q (λ_{1} I - Λ) Q^{T}$ , The eigenvalues are $0, λ_{1} - λ_{2}, \dots, λ_{1} - λ_{n}$ , all $\geq 0$ , so it’s positive semidefinite.
1.
For any $x$ , we have $λ_{1} x^{T} x - x^{T} Sx = x^{T} (λ_{1} I - S) x \geq 0$ by the property of a positive semidefinite matrix $λ_{1} I - S$ .
1.
Since $x^{T} x > 0$ for nonzero $x$ , so we conclude that the maximum value of $\frac{x^{T} Sx}{x^{T} x}$ is $λ_{1}$ .

niuers

2020-03-20 00:00

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