GATE 2025 DA Q.60: solution and answer

Official answer: 100Source: Official answer key, GATE 2025 DA (IIT Roorkee)

1. The data are centred, ${\sum }_{i=1}^n{x}^{(i)}=0$, so the sample mean is $\bar{x}=0$. The covariance matrix (with the $\frac{1}{n}$ convention, which is the one the quantity in the question matches) is $$C=\frac{1}{n}\sum _{i=1}^n{x}^{(i)}{x}^{(i)\top }.$$

2. Expand the required quantity: $$\frac{1}{n}\sum _{i=1}^n{\left(u^{\top }{x}^{(i)}\right)}^2=\frac{1}{n}\sum _{i=1}^n{u}^{\top }{x}^{(i)}{x}^{(i)\top }u=u^{\top }\left(\frac{1}{n}\sum _{i=1}^n{x}^{(i)}{x}^{(i)\top }\right)u=u^{\top }Cu.$$

3. Because the mean is zero, $u^{\top }Cu$ is exactly the variance of the data projected on the unit vector $u$.

4. $C$ is symmetric, so write $C={\sum }_j{\lambda }_j{v}_j{v}_j^{\top }$ with orthonormal eigenvectors $v_j$. For a unit vector $u={\sum }_j{a}_j{v}_j$ with ${\sum }_j{a}_j^2=1$, we get $u^{\top }Cu={\sum }_j{\lambda }_j{a}_j^2\le {\lambda }_{\max }{\sum }_j{a}_j^2={\lambda }_{\max }$. Equality holds when $u$ is the eigenvector of the largest eigenvalue. So the direction of maximum variance gives $u^{\top }Cu={\lambda }_{\max }$.

5. The eigenvalues are ${\lambda }_i=100^{2-i}$, which strictly decrease as $i$ increases: ${\lambda }_1=100^1=100$, ${\lambda }_2=100^0=1$, ${\lambda }_3=100^{-1}=0.01$, and so on. Hence ${\lambda }_{\max }={\lambda }_1=100$.

6. Therefore $$\frac{1}{n}\sum _{i=1}^n{\left(u^{\top }{x}^{(i)}\right)}^2={\lambda }_1=100.$$