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Exercise 3.2.12

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We have ${(Cy)}^{T} S (Cy) = z^{T} Sz$ , where $z = Cy$ , since $y \neq 0$ , and $C$ is invertible, so $C$ has full rank, and $Cy \neq 0$ , so by the definition of positive definite matrix $S$ , we have $z^{T} Sz 0$ .

niuers

2020-03-20 00:00

Exercise 3.2.12

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