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.