GATE 2025 EE Q.12: solution and answer

Official answer: BSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Let $A$ be the real symmetric matrix ($A^T=A$), with $A{\mathbf{v}}_1={\lambda }_1{\mathbf{v}}_1$, $A{\mathbf{v}}_2={\lambda }_2{\mathbf{v}}_2$ and ${\lambda }_1\ne {\lambda }_2$. Eigenvectors are nonzero by definition.

2. Premultiply the second equation by ${\mathbf{v}}_1^T$: ${\mathbf{v}}_1^{T}A{\mathbf{v}}_2={\lambda }_2{\mathbf{v}}_1^T{\mathbf{v}}_2$.

3. Use symmetry: ${\mathbf{v}}_1^{T}A=(A^T{\mathbf{v}}_1{)}^T=(A{\mathbf{v}}_1{)}^T={\lambda }_1{\mathbf{v}}_1^T$. So the same scalar is also ${\mathbf{v}}_1^{T}A{\mathbf{v}}_2={\lambda }_1{\mathbf{v}}_1^T{\mathbf{v}}_2$.

4. Subtract the two results: $$({\lambda }_1-{\lambda }_2){\mathbf{v}}_1^T{\mathbf{v}}_2=0.$$ Since ${\lambda }_1\ne {\lambda }_2$, ${\mathbf{v}}_1^T{\mathbf{v}}_2=0$. The eigenvectors are orthogonal, so (B) is true and (A) is false.

5. (C) would give ${\mathbf{v}}_2=-{\mathbf{v}}_1$ and (D) would give ${\mathbf{v}}_2={\mathbf{v}}_1$. In either case $A{\mathbf{v}}_2={\lambda }_1{\mathbf{v}}_2$ as well as $A{\mathbf{v}}_2={\lambda }_2{\mathbf{v}}_2$, so $({\lambda }_1-{\lambda }_2){\mathbf{v}}_2=0$. That is impossible, because ${\mathbf{v}}_2\ne 0$ and ${\lambda }_1\ne {\lambda }_2$. (Also, ${\mathbf{v}}_1^T{\mathbf{v}}_2=\mp \Vert {\mathbf{v}}_1{\Vert }^2\ne 0$ would contradict the orthogonality just proved.) The answer is (B).