Question:
Let \( D = \{ x^{(1)}, x^{(2)}, \dots, x^{(n)} \} \) be a dataset of \( n \) observations where each \( x^{(i)} \in \mathbb{R}^{100} \). It is given that
\[
\sum_{i=1}^{n} x^{(i)} = 0.
\]
The covariance matrix computed from \( D \) has eigenvalues \( \lambda_i = 100^2 - i \), for \( 1 \leq i \leq 100 \). Let \( u \in \mathbb{R}^{100} \) be the direction of maximum variance with \( u^T u = 1 \). The value of
\[
\frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 = \underline{\hspace{2cm}} \quad \text{(Answer in integer)}
\]
Let \( D = \{ x^{(1)}, x^{(2)}, \dots, x^{(n)} \} \) be a dataset of \( n \) observations where each \( x^{(i)} \in \mathbb{R}^{100} \). It is given that
\[ \sum_{i=1}^{n} x^{(i)} = 0. \] The covariance matrix computed from \( D \) has eigenvalues \( \lambda_i = 100^2 - i \), for \( 1 \leq i \leq 100 \). Let \( u \in \mathbb{R}^{100} \) be the direction of maximum variance with \( u^T u = 1 \). The value of
\[ \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 = \underline{\hspace{2cm}} \quad \text{(Answer in integer)} \]
\[ \sum_{i=1}^{n} x^{(i)} = 0. \] The covariance matrix computed from \( D \) has eigenvalues \( \lambda_i = 100^2 - i \), for \( 1 \leq i \leq 100 \). Let \( u \in \mathbb{R}^{100} \) be the direction of maximum variance with \( u^T u = 1 \). The value of
\[ \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 = \underline{\hspace{2cm}} \quad \text{(Answer in integer)} \]
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In PCA, the first principal component is the direction corresponding to the largest eigenvalue of the covariance matrix. The average squared projection onto this component gives the variance along the direction of maximum variance.
Updated On: Apr 4, 2025
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Solution and Explanation
Step 1: Understanding the given condition.
We are given that the covariance matrix has eigenvalues \( \lambda_i = 100^2 - i \) for \( 1 \leq i \leq 100 \). The direction of maximum variance, \( u \), corresponds to the first principal component, and it is the eigenvector associated with the largest eigenvalue \( \lambda_1 = 100^2 \).
Step 2: Principal Component Analysis.
The expression \( u^T x^{(i)} \) represents the projection of the \( i \)-th data point onto the principal component \( u \). The term \( \left( u^T x^{(i)} \right)^2 \) is the squared value of this projection.
The sum \( \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 \) corresponds to the average of the squared projections of the data points onto the first principal component.
Step 3: Eigenvalue decomposition of the covariance matrix.
The variance along each direction is equal to the corresponding eigenvalue. Since \( u \) is the direction of maximum variance, the variance in the direction of \( u \) is equal to \( \lambda_1 = 100^2 \). Therefore, the average squared projection is:
\[ \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 = \lambda_1 = 100. \]
Thus, the value of \( \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 \) is \( 100 \).
We are given that the covariance matrix has eigenvalues \( \lambda_i = 100^2 - i \) for \( 1 \leq i \leq 100 \). The direction of maximum variance, \( u \), corresponds to the first principal component, and it is the eigenvector associated with the largest eigenvalue \( \lambda_1 = 100^2 \).
Step 2: Principal Component Analysis.
The expression \( u^T x^{(i)} \) represents the projection of the \( i \)-th data point onto the principal component \( u \). The term \( \left( u^T x^{(i)} \right)^2 \) is the squared value of this projection.
The sum \( \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 \) corresponds to the average of the squared projections of the data points onto the first principal component.
Step 3: Eigenvalue decomposition of the covariance matrix.
The variance along each direction is equal to the corresponding eigenvalue. Since \( u \) is the direction of maximum variance, the variance in the direction of \( u \) is equal to \( \lambda_1 = 100^2 \). Therefore, the average squared projection is:
\[ \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 = \lambda_1 = 100. \]
Thus, the value of \( \frac{1}{n} \sum_{i=1}^{n} \left( u^T x^{(i)} \right)^2 \) is \( 100 \).
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