From 172ddce72b0f2cad97e21ba0f441581ed8cc5e4b Mon Sep 17 00:00:00 2001 From: Vladimir Feinberg Date: Mon, 12 Jun 2017 10:44:57 -0400 Subject: [PATCH] Fixed typos in linear algebra problem * Transposition operator should be superscripted * Added convexity justification for critical-point optimum --- linear_algebra.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/linear_algebra.tex b/linear_algebra.tex index 9e3972d..adf2647 100644 --- a/linear_algebra.tex +++ b/linear_algebra.tex @@ -40,11 +40,11 @@ \section*{Solutions} In the last line, we used the fact that $\vu$ is a unit vector to make the simplification $\vu^\top \vu = 1$. -We can minimize this distance by taking the derivative with respect to -$\alpha$ and setting it to zero: +We can globally minimize this distance by taking the derivative with respect to +$\alpha$ and setting it to zero, since the objective is convex and differentiable: \begin{align} & - 2 \vx^\top \vu + 2 \alpha = 0 \\ -\Rightarrow & \alpha = \vx\top \vu. +\Rightarrow & \alpha = \vx^\top \vu. \end{align} Recalling that $\vy = \alpha \vu$, we can conclude that $\vy = \vx^\top \vu \vu$.