← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q3c-i — Step-by-Step Solution

8 marks · Section A

Symmetric and Skew-Symmetric Matrices · Linear Algebra · Read the full method →

Question

Prove that eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal.

Technique

Standard “compute $v_2^T A v_1$ two ways” argument; the two computations give $\lambda_1=\lambda_2$ or orthogonality.

Solution

Setup. $A$ real symmetric, so $A^T=A$. Let $\lambda_1\ne\lambda_2$ be distinct eigenvalues with eigenvectors $v_1,v_2$:

To show: $v_1^T v_2=0$.

Step 1 — Take inner product

$v_2^T(Av_1)=v_2^T(\lambda_1 v_1)=\lambda_1(v_2^T v_1)$.

Step 2 — Use symmetry

$v_2^T A v_1=(Av_2)^T v_1$ (transposing the scalar $v_2^T A v_1$, using $A^T=A$): $=(\lambda_2 v_2)^T v_1=\lambda_2(v_2^T v_1)$.

Step 3 — Equate

$\lambda_1(v_2^T v_1)=\lambda_2(v_2^T v_1)$, $(\lambda_1-\lambda_2)(v_2^T v_1)=0$.

Since $\lambda_1\ne\lambda_2$, we have $\lambda_1-\lambda_2\ne 0$, so $v_2^T v_1=0$.

Answer