GATE 2026 DA Q.65: solution and answer

Official answer: 1Source: Official answer key, GATE 2026 DA (IIT Guwahati)

1. Write $P=\frac{1}{n}11^T$, so $A=I_n-P$. Since $1^{T}1=n$, we get $P^2=\frac{1}{n^2}1(1^{T}1)1^T=P$, and $P^T=P$. So $P$ is an orthogonal projection of rank 1.

2. Then $A^T=A$ and $A^2=I-2P+P^2=I-P=A$. So $A$ is symmetric and idempotent, i.e. the orthogonal projection onto the subspace perpendicular to $1$.

3. An idempotent matrix has eigenvalues only 0 or 1. Here $A1=1-\frac{1}{n}1(1^{T}1)=1-1=0$, so $0$ is an eigenvalue. The other $n-1$ eigenvalues are $1$, because $\mathrm{rank}(A)=n-1$ (for $n\ge 2$).

4. For a real symmetric matrix, ${\max }_{x^{T}x=1}{x}^{T}Ax={\lambda }_{\max }(A)$ (Rayleigh quotient), so the maximum is $1$.

5. Direct check: for any unit $x$, $x^{T}Ax=x^{T}x-\frac{1}{n}(1^{T}x{)}^2=1-\frac{(1^{T}x{)}^2}{n}\le 1$, with equality exactly when $1^{T}x=0$. Such unit vectors exist for $n\ge 2$, for example $x=\frac{1}{\sqrt{2}}(1,-1,0,\dots ,0{)}^T$.

6. Hence ${\max }_S{x}^{T}Ax=1$.