Let D = \x^(1), , x^(n)\ be a dataset of n observations where each x^(i) ∈ R^100. It is given that _i=1^n x^(i) = 0. The covariance matrix computed from D has eigenvalues _i = 100^2-i, 1 ≤ i ≤ 100. Let u ∈ R^100 be the direction of maximum variance with u^u = 1. The value of 1/n _i=1^n ≤ft(u^x^(i))^2 = \_\_\_\_ (Answer in integer)

GATE DA 2025 Q60: PCA: variance along the top principal direction

Machine Learning
2 marks
NAT
No negative marking

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)

Try it: numerical answer: type the value

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Numerical answer. No negative marking.

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Answer: 100 (official GATE 2025 final answer key)

Concept used: The variance along the first principal direction equals the largest eigenvalue of the covariance matrix.

First step: With zero-mean data, $\frac{1}{n}\sum (u^{\top}x^{(i)})^2 = u^{\top}\Sigma u$ , which is maximised by the top eigenvector. List the eigenvalues: $\lambda_1 = 100^{1}$ , $\lambda_2 = 100^{0}$ , and so on.

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Source: official GATE 2025 DA question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key

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GATE DA 2025 Q60: PCA: variance along the top principal direction

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