# Hermitian Matrices

## Table of Contents

## 1 Definition

We say a complex square matrix \( A \) is **Hermitian** if it is equal to its own **conjugate transpose**, ie \( \forall i, j \) we have that:

\[ A_{ij} = \overline{A_{ji}} \]

Alternate names for the conjugate transpose of a matrix include the **adjoint** or **transjugate**, and may be referred to by any of the following symbols:

\[ A^H \equiv A^* \equiv \overline{A^{T}} \]

We can view Hermitian matrices as an extension of real symmetric matrices as they share many of the same properties.

## 2 The Spectral Theorem

We give the Spectral Theorem in the complex case: *If an \( n \times n \) matrix \( A \) is Hermitian, then all eigenvalues of \( A \) are real and there exists a orthonormal basis for \( \mathbb{C}^n \) consisting of the eigenvectors of \( A \).*

Since \( A \) has \( n \) orthogonal (and thus linearly independent) eigenvectors, we arrive at the corollary that \( A \) is **unitarily** diagonalizable.

This theorem makes many claims, each of which we will prove in turn. The first two of these claims are straightforward to show, whilst the last is less so.

### 2.1 All Eigenvalues of \( A \) are Real

Let \( \lambda \) be an eigenvalue of \( A \), with corresponding eigenvector \( v \) with entries given by \( a_k + ib_k \). Then starting with \( Av = \lambda v \) :

\begin{align*} \left(v^*\right)^T Av &= \left(v^*\right)^T \lambda v \\ &= \lambda \left(v^*\right)^T v \\ &= \lambda [a_1^2 + b_1^2 + ... + a_n^2 + b_n^2] \text{ (a 1} \times \text{1 matrix)} \end{align*}Now taking the conjugate transpose of both sides, (note the LHS is invariant under this operation) gives:

\begin{array}{r l l} &(\left(v^*\right)^T Av)^* &= (\lambda [a_1^2 + b_1^2 + ... + a_n^2 + b_n^2])^* \\ \implies& \left(v^*\right)^T Av &= \lambda^* [a_1^2 + b_1^2 + ... + a_n^2 + b_n^2] \\ \implies& \lambda [a_1^2 + b_1^2 + ... + a_n^2 + b_n^2] &= \lambda^* [a_1^2 + b_1^2 + ... + a_n^2 + b_n^2] \\ \implies& \lambda &= \lambda^* \\ \implies& \lambda \in \mathbb{R} \end{array}### 2.2 All Eigenvectors of \( A \) are Orthogonal

Let \( v_1 \) and \( v_2 \) be eigenvectors of \( A \) corresponding to the distinct eigenvalues \( \lambda_1 \) and \( \lambda_2 \). Consider the product \( (Av_1)^*v_2 \). First of all we have that:

\begin{align*} (Av_1)^*v_2 &= v_1^*A^*v_2 \\ &= v_1^*Av_2 \\ &= v_1^*\lambda_2 v_2 \\ &= \lambda_2 v_1^* v_2 \end{align*}On the other hand, simplifying inside the bracket first gives:

\begin{align*} (Av_1)^*v_2 &= (\lambda_1 v_1)^*v_2 \\ &= \lambda_1^* v_1^*v_2\\ &= \lambda_1 v_1^* v_2 \end{align*}Equating these gives \( (\lambda_1 - \lambda_2)v_1^* v_2 = 0 \) from which we must conclude that \( v_1^* \) and \( v_2 \) are orthogonal since the two eigenvalues are distinct.