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world:pca [2022/11/12 13:50]
rdouc [Proof]
world:pca [2022/11/13 18:38] (current)
rdouc [Proof]
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 ====== Statement ====== ====== Statement ======
  
-Let $(X_i)_{1\leq i \leq n}$ be $n$ observations in $\rset^d$. Define $\Hset_p$ the set of all the subspaces of $\rset^d$ of dimension $p$ and write $\projorth{H}X$ the orthogonal projection of a vector $X \in \rset^d$ on a subspace $\Hset$. +Let $(X_i)_{1\leq i \leq n}$ be $n$ observations in $\rset^d$. Define $\Hset_p$ the set of all the subspaces of $\rset^d$ of dimension $p$ and write $\projorth{H}X$ the orthogonal projection of a vector $X \in \rset^d$ on a subspace $H$. 
  
 In Principal Component Analysis (PCA), we consider the optimisation problem: ​ In Principal Component Analysis (PCA), we consider the optimisation problem: ​
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     &=w^T S_n w=w^T \lr{\sum_{j=1}^d \lambda_j ​ w_j w_j^T} w=\sum_{j=1}^d \lambda_j (w^T w_j)^2 \label{eq:​dim1}     &=w^T S_n w=w^T \lr{\sum_{j=1}^d \lambda_j ​ w_j w_j^T} w=\sum_{j=1}^d \lambda_j (w^T w_j)^2 \label{eq:​dim1}
 \end{align} \end{align}
 +Therefore, we have 
 \begin{align*} \begin{align*}
-    &\leq \lambda_1 \sum_{j=1}^d (w^T w_j)^2=\lambda_1 \norm{w}^2=\lambda_1 ​+    ​\sum_{i=1}^n \norm{\projorth{\rset w}X_i}^2 ​&\leq \lambda_1 \sum_{j=1}^d (w^T w_j)^2=\lambda_1 \norm{w}^2=\lambda_1 ​
 \end{align*} \end{align*}
-Since $\sum_{i=1}^n \norm{\projorth{\rset ​w_1}X_i}^2=w_1^T S_n w_1=\lambda_1$. Therefore \eqref{eq:​vp} holds true for $p=1$. ​+Note that from \eqref{eq:​dim1},​ we have for any eigenvector ​$w_k$ of $S_n$, ​  
 +\begin{equation} \label{eq:​eigenvector} 
 +\sum_{i=1}^n \norm{\projorth{\rset ​w_k}X_i}^2=w_k^T S_n w_k=\lambda_k. 
 +\end{equation} 
 +In particular, $\sum_{i=1}^n \norm{\projorth{\rset ​w_1}X_i}^2=\lambda_1$. Therefore \eqref{eq:​vp} holds true for $p=1$. ​
  
-Assume now that \eqref{eq:​vp} hold true for some $p \in [1:d-1]$ and let $H \in \Hset_p$. Then, denote by $G=\mathrm{Span}(w_1,​\ldots,​w_p)^\perp$. Since $\mathrm{dim} (G)=d-p$ and $\mathrm{dim}(H)=p+1$,​ we must have $G \cap H \notin \{0\}$ (Otherwise the subspace $G+H$ would be of dimension $n-p+p+1=n+1$ which is not possible). Let $w_0$ a unitary vector of $G \cap H$. Then, we have the decomposition $H=\rset w_0 \stackrel{\perp}{+} H_0$ where $H_0$ is of dimension $p$. Then, +Assume now that \eqref{eq:​vp} hold true for some $p \in [1:d-1]$ and let $H \in \Hset_{p+1}$. Then, denote by $G=\mathrm{Span}(w_1,​\ldots,​w_p)^\perp$. Since $\mathrm{dim} (G)=d-p$ and $\mathrm{dim}(H)=p+1$,​ we must have $G \cap H \notin \{0\}$ (Otherwise the subspace $G+H$ would be of dimension $d-p+p+1=d+1$ which is not possible). Let $w_0$ a unitary vector of $G \cap H$. Then, we have the decomposition $H=\rset w_0 \stackrel{\perp}{+} H_0$ where $H_0$ is of dimension $p$. Then, applying \eqref{eq:​dim1}
 \begin{align*} \begin{align*}
     \sum_{i=1}^n \norm{\projorth{H}X_i}^2&​= \sum_{i=1}^n \norm{\projorth{\rset w_0}X_i}^2+\norm{\projorth{H_0}X_i}^2\\     \sum_{i=1}^n \norm{\projorth{H}X_i}^2&​= \sum_{i=1}^n \norm{\projorth{\rset w_0}X_i}^2+\norm{\projorth{H_0}X_i}^2\\
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     &​=\sum_{j=p+1}^d \lambda_j (w_0^T w_j)^2 +\sum_{i=1}^n \norm{\projorth{H_0}X_i}^2\\     &​=\sum_{j=p+1}^d \lambda_j (w_0^T w_j)^2 +\sum_{i=1}^n \norm{\projorth{H_0}X_i}^2\\
 \end{align*} \end{align*}
-where we used that $w_0 \in G=\mathrm{Span}(w_1,​\ldots,​w_p)^\perp$. Applying ​now \eqref{eq:dim1and the induction assumption+where we used that $w_0 \in G=\mathrm{Span}(w_1,​\ldots,​w_p)^\perp$. Applying ​the induction assumption and then \eqref{eq:eigenvector},
 \begin{align*} \begin{align*}
     \sum_{i=1}^n \norm{\projorth{H}X_i}^2&​ \leq \lambda_{p+1} \sum_{j=p+1}^d (w_0^T w_j)^2 +\sum_{i=1}^n \norm{\projorth{\mathrm{Span}(w_1,​\ldots,​w_p)}X_i}^2\\     \sum_{i=1}^n \norm{\projorth{H}X_i}^2&​ \leq \lambda_{p+1} \sum_{j=p+1}^d (w_0^T w_j)^2 +\sum_{i=1}^n \norm{\projorth{\mathrm{Span}(w_1,​\ldots,​w_p)}X_i}^2\\
world/pca.1668257430.txt.gz · Last modified: 2022/11/12 13:50 by rdouc