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fixed pca issue

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1 parent 14af030 commit fc13bafe1960d458ef45365251583d35b61a1a9e @hal3 committed Dec 6, 2015
Showing with 7 additions and 6 deletions.
  1. +7 −6 book/unsup.tex
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@@ -286,9 +286,10 @@ \section{Linear Dimensionality Reduction}
\dots, \vx_N$ and you're looking for a vector $\vec u$ that points in
the direction of maximal variance. You can compute this by projecting
each point onto $\vec u$ and looking at the variance of the result.
-In order for the projection to make sense, you need to constrain $\norm{\vec
-u}^2 = 1$. In this case, the projections are $\dotp{\vx_1, \vec u},
-\dots, \dotp{\vx_N, \vec u}$. Call these values $p_1, \dots, p_N$.
+In order for the projection to make sense, you need to constrain
+$\norm{\vec u}^2 = 1$. In this case, the projections are
+$\dotp{\vx_1}{\vec u}, \dotp{\vx_2}{\vec u}, \dots, \dotp{\vx_N}{\vec
+ u}$. Call these values $p_1, \dots, p_N$.
The goal is to compute the variance of the $\{p_n\}$s and then choose
$\vec u$ to maximize this variance. To compute the variance, you
@@ -362,12 +363,12 @@ \section{Linear Dimensionality Reduction}
\begin{align}
\cL(\vec v, \la_1, \la_2) &=
\norm{ \mat X \vec v }^2
- - \la_1 \left( \norm{\vec u}^2 - 1 \right)
+ - \la_1 \left( \norm{\vec v}^2 - 1 \right)
- \la_2 \dotp{\vec u}{\vec v}\\
\grad_{\vec u} \cL &=
- 2 \mat X\T\mat X \vec v - 2 \la_1 \vec v - 2 \la_2 \vec u\\
+ 2 \mat X\T\mat X \vec v - 2 \la_1 \vec v - \la_2 \vec u\\
\Longrightarrow & \quad\la_1 \vec v = \left(\mat X\T\mat X\right)\vec v
-- \la_2 \vec u
+- \frac {\la_2} 2 \vec u
\end{align}
%
However, you know that $\vec u$ is the first eigenvector of $\mat

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