Hermitian matrix

From Mathematics wiki

A matrix that is equal to its Hermitian adjoint, i.e., is self-adjoint:

$H^{\dagger} = H^{*T} = H\,$ .

A Hermitian matrix is in some sense the opposite of a anti-Hermitian matrix and is the analog of a purely-real number. It is also the complex extension of the concept of a symmetric matrix. Note that if $H\,$ is Hermitian then $i H\,$ is anti-Hermitian.

Any complex matrix $X\,$ may be decomposed into its Hermitian and anti-Hermitian parts, namely $X + X^\dagger\,$ and $X - X^\dagger\,$ , respectively.

Eigenvalues and Eigenvectors

Theorem:

Eigenvalues of a Hermitian matrix are real

Proof:

Suppose $\mathbf{v}\,$ is a (non-zero) eigenvector of $H\,$ with eigenvalue $\lambda\,$ ,

$H \mathbf{v} = \lambda \mathbf{v}\,$ .

Then

$\lambda^* \mathbf{v}^\dagger \mathbf{v}\,$	$=(H \mathbf{v})^\dagger \mathbf{v}\,$ ,
	$=\mathbf{v}^\dagger H^\dagger\mathbf{v}\,$ ,
	$=\mathbf{v}^\dagger H \mathbf{v}\,$ ,
	$=\lambda \mathbf{v}^\dagger \mathbf{v}\,$ .

Thus either $\mathbf{v} = 0\,$ , contrary to assumption, or $\lambda^* = \lambda\,$ .■

Theorem:

Eigenvectors corresponding to distinct eigenvalues are orthogonal

Proof:

Suppose $\mathbf{v}_1\,$ and $\mathbf{v}_2\,$ are eigenvectors of $A\,$ with corresponding eigenvalues $\lambda_1\,$ and $\lambda_2\,$ respectively. Then

$\lambda_2 \mathbf{v}_2^\dagger \mathbf{v}_1\,$	$= \mathbf{v}_2^\dagger H^\dagger \mathbf{v}_1\,$ ,
	$= \mathbf{v}_2^\dagger H \mathbf{v}_1\,$ ,
	$= \lambda_1 \mathbf{v}_2^\dagger \mathbf{v}_1\,$ ,

so $(\lambda_2 - \lambda_1)\mathbf{v}_2^\dagger \mathbf{v}_1=0\,$ . Thus either $\lambda_1 = \lambda_2\,$ , contrary to assumption, or $\mathbf{v}_2^\dagger \mathbf{v}_1 = 0\,$ . ■

It may happen that more than one eigenvector correspond to a single eigenvalue. Such a set spans a subspace of $V^n\,$ called an eigenspace. Having found these eigenvectors, we may orthogonalize this set, and since any linear combination of these vectors has the same eigenvalue, the resulting set again consists of eigenvectors of $H\,$ .

Measure

We can express the measure $[dH]\,$ on the space of Hermitian matrices in terms of the eigenvalues $\lambda_i\,$ through a change of variables. Write $H = \Omega \Lambda \Omega^\dagger\,$ where $\Lambda = \operatorname{diag}(\lambda_1, \dots,\lambda_n)\,$ is diagonal and $\Omega \in U(n) / U(1)^n\,$ is unitary. Differentiate:

$dH = d\Omega \Lambda \Omega + \Omega d\Lambda \Omega^\dagger + \Omega \Lambda d\Omega^\dagger = d\Omega \Lambda \Omega + \Omega d\Lambda \Omega^\dagger - \Omega \Lambda \Omega^\dagger d\Omega\,$ ,

whence

$\Omega^\dagger dH \Omega = d\Lambda + [\Omega^\dagger d\Omega, \Lambda]\,$ .

Now $d\omega = \Omega^\dagger d\Omega\,$ is an infinitesimal element on the Lie group U(n) near the identity, i.e., an element of the Lie algebra u(n) and therefore an anti-Hermitian matrix. Now,

$(\Omega^\dagger dH \Omega)_{ij} = d\lambda_i \delta_{ij} + d\omega_{ij} (\lambda_i - \lambda_j)\,$ ,

so that we can read the metric off of the Frobenius norm which is invariant under unitary transformations:

$\left\Vert \delta H \right\Vert ^2 = \operatorname{tr} \left( \delta H \delta H^\dagger\right)\,$ .

Thus,

$ds^2 = \sum_i d\lambda_i^2 + 2\sum_{1 \leq i < j \leq n} (\lambda_i - \lambda_j)^2 d\omega_{ij} d\omega_{ij}^*\,$ ,

or, more transparently,

$ds^2 = \sum_i d\lambda_i^2 + 2\sum_{1 \leq i < j \leq n} (\lambda_i - \lambda_j)^2 \left[ (d\,\operatorname{Re}\, \omega_{ij})^2 + (d\,\operatorname{Im}\, \omega_{ij})^2 \right]\,$ .