Welcome to Randal Douc's wiki

You've loaded an old revision of the document! If you save it, you will create a new version with this data.

====== 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 $H$.

In Principal Component Analysis (PCA), we consider the optimisation problem: 
\begin{align*}
    V_p&= \argmin_{H \in \Hset_p} \sum_{i=1}^n \norm{X_i - \projorth{H}X_i}^2 \\
    &=\argmin_{H \in \Hset_p} \sum_{i=1}^n \norm{X_i}^2 - \norm{\projorth{H}X_i}^2\\
    &=\argmax_{H \in \Hset_p} \sum_{i=1}^n \norm{\projorth{H}X_i}^2
\end{align*}

Define $S_n= \sum_{i=1}^n X_i X_i^T$ and since $S_n$ is a symmetric and positive matrix with real entries, it is diagonizable in an orthonormal basis $(w_j)_{1\leq j \leq d}$ with eigenvalues $(\lambda_j)_{1\leq j \leq d}$ ranked in a decreasing order, that is, $\lambda_1 \geq \ldots \geq \lambda_n \geq 0$. Defining $D={\mathrm {Diag}} ((\lambda_j)_{1\leq j \leq d})$ and $U=[w_1,\ldots,w_d]$ we have $S_n = U D U^T=\sum_{j=1}^d \lambda_j w_j w_j^T$.

<note tip>    For any $p \in  [1:d]$, we have 
\begin{equation} \label{eq:vp}
        V_p=\mathrm{Span}(w_1,\ldots,w_p)
\end{equation} 
</note>
===== Proof =====

The proof is by induction. We start with $p=1$. For any unitary vector $w\in \rset^d$, 
\begin{align}
    \sum_{i=1}^n \norm{\projorth{\rset w}X_i}^2&=\sum_{i=1}^n (X_i^T w)^2=\sum_{i=1}^n w^T X_i X_i^T w=w^T (\sum_{i=1}^n X_i X_i^T) w \nonumber\\
    &=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}
Therefore, we have 
\begin{align*}
    \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*}
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 $p=1$.

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, applying \eqref{eq:dim1}
\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_{j=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*}
where we used that $w_0 \in G=\mathrm{Span}(w_1,\ldots,w_p)^\perp$. Applying the induction assumption and then \eqref{eq:dim1},
\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\\
    & \leq \lambda_{p+1} \norm{w_0}^2+\sum_{i=1}^n \norm{\projorth{\mathrm{Span}(w_1,\ldots,w_p)}X_i}^2\\
    & = \lambda_{p+1} +\sum_{i=1}^n \norm{\projorth{\mathrm{Span}(w_1,\ldots,w_p)}X_i}^2\\
    & = \sum_{i=1}^n \norm{\projorth{\rset w_{p+1}}X_i}^2 +\sum_{i=1}^n \norm{\projorth{\mathrm{Span}(w_1,\ldots,w_p)}X_i}^2\\
    & = \sum_{i=1}^n \lr{\norm{\projorth{\rset w_{p+1}}X_i}^2 +\norm{\projorth{\mathrm{Span}(w_1,\ldots,w_p)}X_i}^2}\\
    & = \sum_{i=1}^n \norm{\projorth{\mathrm{Span}(w_1,\ldots,w_{p+1})}X_i}^2\\
\end{align*}  
which concludes the proof by an induction argument.

Please fill all the letters into the box to prove you're human. S V P I E Please keep this field empty:

Edit summary

Welcome to Randal Douc's wiki

User Tools

Site Tools

Page Tools