Merge branch 'master' of skozl.com:e4-pattern

1 files changed, 46 insertions, 38 deletions

diff --git a/report/paper.md b/report/paper.md
index dc22600..961937b 100755
--- a/report/paper.md
+++ b/report/paper.md
@@ -70,46 +70,11 @@ two computation techniques used shows that the difference
 is very small (due to rounding
 of the np.eigh function when calculating the eigenvalues
 and eigenvectors of the matrices A\textsuperscript{T}A (NxN) and AA\textsuperscript{T}
-(DxD)).
-
-The first ten biggest eigenvalues obtained with each method
-are shown in table \ref{tab:eigen}.
-
-\begin{table}[ht]
-\centering
-\begin{tabular}[t]{cc}
-PCA &Fast PCA\\
-2.9755E+05 &2.9828E+05\\
-1.4873E+05 &1.4856E+05\\
-1.2286E+05 &1.2259E+05\\
-7.5084E+04 &7.4950E+04\\
-6.2575E+04 &6.2428E+04\\
-4.7024E+04 &4.6921E+04\\
-3.7118E+04 &3.7030E+04\\
-3.2101E+04 &3.2046E+04\\
-2.7871E+04 &2.7814E+04\\
-2.4396E+04 &2.4339E+04\\
-\end{tabular}
-\caption{Comparison of eigenvalues obtain with the two computation methods}
-\label{tab:eigen}
-\end{table}
+(DxD)). The first ten biggest eigenvalues obtained with each method
+are shown in Appendix, table \ref{tab:eigen}.
 
 It can be proven that the eigenvalues obtained are mathematically the same [@lecture-notes],
-and the there is a relation between the eigenvectors obtained: 
-
-Computing the eigenvectors **u\textsubscript{i}** for the DxD matrix AA\textsuperscript{T} 
-we obtain a very large matrix. The computation process can get very expensive when $D \gg N$.
-
-For such reason we compute the eigenvectors **v\textsubscript{i}** of the NxN
-matrix A\textsuperscript{T}A. From the computation it follows that $A\textsuperscript{T}A\boldsymbol{v\textsubscript{i}} = \lambda \textsubscript{i}\boldsymbol{v\textsubscript{i}}$.
-
-Multiplying both sides by A we obtain: 
-
-$$ AA\textsuperscript{T}A\boldsymbol{v\textsubscript{i}} = \lambda \textsubscript{i}A\boldsymbol{v\textsubscript{i}} \rightarrow SA\boldsymbol{v\textsubscript{i}} = \lambda \textsubscript{i}A\boldsymbol{v\textsubscript{i}} $$
-
-We know that $S\boldsymbol{u\textsubscript{i}} = \lambda \textsubscript{i}\boldsymbol{u\textsubscript{i}}$.
- 
-From here it follows that AA\textsuperscript{T} and A\textsuperscript{T}A have the same eigenvalues and their eigenvectors follow the relationship $\boldsymbol{u\textsubscript{i}} = A\boldsymbol{v\textsubscript{i}}$ 
+and the there is a relation between the eigenvectors obtained: $\boldsymbol{u\textsubscript{i}} = A\boldsymbol{v\textsubscript{i}}$. (*Proof in the appendix*). 
 
 It can be noticed that we effectively don't lose any data calculating the eigenvectors
 for PCA with the second method. The main advantages of it are in terms of speed,
@@ -504,3 +469,46 @@ Seed & Individual$(M=120)$ & Bag + Feature Ens.$(M=60+95)$\\ \hline
 \end{table}
 
 # References
+
+#Appendix
+
+##Eigenvectors and Eigenvalues in fast PCA
+
+**Table showing eigenvalues obtained with each method**
+
+\begin{table}[ht]
+\centering
+\begin{tabular}[t]{cc}
+PCA &Fast PCA\\
+2.9755E+05 &2.9828E+05\\
+1.4873E+05 &1.4856E+05\\
+1.2286E+05 &1.2259E+05\\
+7.5084E+04 &7.4950E+04\\
+6.2575E+04 &6.2428E+04\\
+4.7024E+04 &4.6921E+04\\
+3.7118E+04 &3.7030E+04\\
+3.2101E+04 &3.2046E+04\\
+2.7871E+04 &2.7814E+04\\
+2.4396E+04 &2.4339E+04\\
+\end{tabular}
+\caption{Comparison of eigenvalues obtain with the two computation methods}
+\label{tab:eigen}
+\end{table}
+
+**Proof of relationship between eigenvalues and eigenvectors in the different methods**
+
+Computing the eigenvectors **u\textsubscript{i}** for the DxD matrix AA\textsuperscript{T} 
+we obtain a very large matrix. The computation process can get very expensive when $D \gg N$.
+
+For such reason we compute the eigenvectors **v\textsubscript{i}** of the NxN
+matrix A\textsuperscript{T}A. From the computation it follows that $A\textsuperscript{T}A\boldsymbol{v\textsubscript{i}} = \lambda \textsubscript{i}\boldsymbol{v\textsubscript{i}}$.
+
+Multiplying both sides by A we obtain: 
+
+$$ AA\textsuperscript{T}A\boldsymbol{v\textsubscript{i}} = \lambda \textsubscript{i}A\boldsymbol{v\textsubscript{i}} \rightarrow SA\boldsymbol{v\textsubscript{i}} = \lambda \textsubscript{i}A\boldsymbol{v\textsubscript{i}} $$
+
+We know that $S\boldsymbol{u\textsubscript{i}} = \lambda \textsubscript{i}\boldsymbol{u\textsubscript{i}}$.
+ 
+From here it follows that AA\textsuperscript{T} and A\textsuperscript{T}A have the same eigenvalues and their eigenvectors follow the relationship $\boldsymbol{u\textsubscript{i}} = A\boldsymbol{v\textsubscript{i}}$ 
+
+