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Linear Algebra and Learning from Data

Exercise 6.1.9

Answers

(a)
$f (\frac{1}{2}, \frac{1}{2}) \leq 0$ because $f (x, y)$ is smooth and convex, and $f (\frac{1}{2}, \frac{1}{2}) = f (\frac{1}{2} (0, 1) + \frac{1}{2} (1, 0)) \leq \frac{1}{2} f (0, 1) + \frac{1}{2} f (1, 0) = 0$
(a)
The Hessian matrix $H = [\begin{matrix} a & b \\ b & c \end{matrix}]$ should be positive semi-definite. so we have $a \geq 0$ , and $ac - b^{2} \geq 0$ .

niuers

2020-03-20 00:00

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