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Exercise 1.11.2

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The length squared of $v - \frac{v^{t} w}{w^{T} w} w$ is :

\begin{array}{l} {(v - \frac{v^{T} w}{w^{T} w} w)}^{T} (v - \frac{v^{T} w}{w^{T} w} w) & = v^{T} v - 2 \frac{v^{T} w v^{T} w}{w^{T} w} + {(\frac{v^{T} w}{w^{T} w})}^{2} w^{T} w \\ = v^{T} v - \frac{| v^{T} w |^{2}}{w^{T} w} \end{array}

This is greater than or equal to zero, so we have $v^{T} v - \frac{| v^{T} w |^{2}}{w^{T} w} \geq 0$ , $| v |_{2}^{2} | w |_{2}^{2} \geq | v^{T} w |^{2}$ , i.e. $| v |_{2} | w |_{2} \geq | v^{T} w |$

niuers

2020-03-20 00:00

Exercise 1.11.2

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