GATE DA 2025 | Question 60
Let \(D = \{x^{(1)}, \ldots, x^{(n)}\}\) 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 \(\lambda_{i} = 100^{2-i}\), \(1 \le i \le 100\). Let \(u \in \mathbb{R}^{100}\) be the direction of maximum variance with \(u^{\top}u = 1\).
The value of
\[ \frac{1}{n}\sum_{i=1}^{n} \left(u^{\top}x^{(i)}\right)^{2} = \]
________ (Answer in integer)
Explanation
Answer: 100
Since \(\sum_{i} x^{(i)} = 0\), the data are already centred, so the covariance matrix is \(C = \frac{1}{n}\sum_{i=1}^{n} x^{(i)}x^{(i)\top}\). Then
\[ \frac{1}{n}\sum_{i=1}^{n} \left(u^{\top}x^{(i)}\right)^{2} = \frac{1}{n}\sum_{i=1}^{n} u^{\top}x^{(i)}x^{(i)\top}u = u^{\top}Cu, \]
the variance of the data along \(u\). Over unit vectors, \(u^{\top}Cu\) is largest for the top eigenvector of \(C\) (the first principal component), and that largest value is the top eigenvalue.
The eigenvalues \(100^{2-i}\) decrease with \(i\): \(100, 1, 0.01, \ldots\). So the value is \(\lambda_{1} = 100^{1} = 100\).
Answer: 100