← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q2b-ii — Step-by-Step Solution

8 marks · Section A

Hermitian and skew-Hermitian matrices · Linear Algebra · asked 3× in 13 yrs · Read the full method →

Question

Prove that eigenvalues of a Hermitian matrix are all real.

Technique

Sandwich $\mathbf x^{*}A\mathbf x$; Hermiticity makes this scalar real, forcing $\lambda=\overline\lambda$.

Solution

Step 1 — Setup

Let $A$ be Hermitian, i.e. $A^{*}=A$ where $A^{*}=\overline{A}^{\top}$ is the conjugate transpose. Let $\lambda$ be an eigenvalue with eigenvector $\mathbf x\ne\mathbf 0$:

Step 2 — Form the scalar $\mathbf x^{*}A\mathbf x$

Pre-multiply by $\mathbf x^{*}$:

Step 3 — Take the conjugate transpose of the scalar

$\mathbf x^{*}A\mathbf x$ is a $1\times1$ matrix (a scalar), so it equals its own conjugate transpose:

using $A^{*}=A$. Hence $\mathbf x^{*}A\mathbf x$ is real (equal to its own conjugate).

Step 4 — Conclude $\lambda$ is real

The conjugate of (1) reads $\overline{\mathbf x^{*}A\mathbf x}=\overline{\lambda}\,\|\mathbf x\|^2$. Since the left side equals $\mathbf x^{*}A\mathbf x=\lambda\|\mathbf x\|^2$ (Step 3 showed it is real, so it equals its own conjugate), we get

As $\mathbf x\ne\mathbf 0$, $\|\mathbf x\|^2>0$, so $\lambda=\overline{\lambda}$, i.e.

Answer