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In mathematics, an Hermitian matrix (or self-adjoint matrix) is a square matrix with complex entries that is equal to its own conjugate transpose – that is, the element in the i-th row and j-th column is equal to the complex conjugate of the element in the j-th row and i-th column, for all indices i and j:

$$a_{ij} = \overline{a_{ji}}\,.$$

If the conjugate transpose of a matrix A is denoted by $$A^\dagger$$, then the Hermitian property can be written concisely as

$$A = A^\dagger\,.$$

Hermitian matrices can be understood as the complex extension of real symmetric matrices.

Hermitian matrices are named after Charles Hermite, who demonstrated in 1855 that matrices of this form share a property with real symmetric matrices of having eigenvalues always real.

Examples

For example,

$$\begin{bmatrix}3&2+i\\ 2-i&1\end{bmatrix}$$

Well-known families of Pauli matrices, Gell-Mann matrices and various generalizations are Hermitian. In theoretical physics such Hermitian matrices usually are multiplied by imaginary coefficients,[1][2] which results in skew-Hermitian matrices (see below).
Properties

The entries on the main diagonal (top left to bottom right) of any Hermitian matrix are necessarily real. A matrix that has only real entries is Hermitian if and only if it is a symmetric matrix, i.e., if it is symmetric with respect to the main diagonal. A real and symmetric matrix is simply a special case of a Hermitian matrix.

Every Hermitian matrix is a normal matrix, and the finite-dimensional spectral theorem applies. It says that any Hermitian matrix can be diagonalized by a unitary matrix, and that the resulting diagonal matrix has only real entries. This implies that all eigenvalues of a Hermitian matrix A are real, and that A has n linearly independent eigenvectors. Moreover, it is possible to find an orthonormal basis of Cn consisting of n eigenvectors of A.

The sum of any two Hermitian matrices is Hermitian, and the inverse of an invertible Hermitian matrix is Hermitian as well. However, the product of two Hermitian matrices A and B will only be Hermitian if they commute, i.e., if AB = BA. Thus An is Hermitian if A is Hermitian and n is an integer.

The Hermitian complex n-by-n matrices do not form a vector space over the complex numbers, since the identity matrix $$I_n$$ is Hermitian, but $$i(I_n)$$ is not. However the complex Hermitian matrices do form a vector space over the real numbers. In the 2n2 R dimensional vector space of complex n×n matrices, the complex Hermitian matrices form a subspace of dimension n2. If Ejk denotes the n-by-n matrix with a 1 in the j,k position and zeros elsewhere, a basis can be described as follows:

$$\; E_{jj} for 1\leq j\leq n$$ (n matrices)

together with the set of matrices of the form

$$\; E_{jk}+E_{kj} for 1\leq j<k\leq n ((n2−n)/2 matrices)$$

and the matrices

$$\; i(E_{jk}-E_{kj}) for 1\leq j<k\leq n ((n2−n)/2$$ matrices)

where i denotes the complex number \sqrt{-1}, known as the imaginary unit.

If n orthonormal eigenvectors u_1,\dots,u_n of a Hermitian matrix are chosen and written as the columns of the matrix U, then one eigendecomposition of A is $$A = U \Lambda U^\dagger$$ where $$U U^\dagger = I=U^\dagger U$$ and therefore

$$A = \sum _j \lambda_j u_j u_j ^\dagger ,$$

where \lambda_j are the eigenvalues on the diagonal of the diagonal matrix $$\; \Lambda .$$

Additional facts related to Hermitian matrices include:

The sum of a square matrix and its conjugate transpose $$(C + C^{\dagger})$$ is Hermitian.
The difference of a square matrix and its conjugate transpose $$(C - C^{\dagger})$$ is skew-Hermitian (also called antihermitian).
This implies that commutator of two Hermitian matrices is skew-Hermitian.
An arbitrary square matrix C can be written as the sum of a Hermitian matrix A and a skew-Hermitian matrix B:

$$C = A+B \quad\mbox{with}\quad A = \frac{1}{2}(C + C^{\dagger}) \quad\mbox{and}\quad B = \frac{1}{2}(C - C^{\dagger}).$$

The determinant of a Hermitian matrix is real:

Proof: $$\det(A) = \det(A^\mathrm{T})\quad \Rightarrow \quad \det(A^\dagger) = \det(A)^*$$
Therefore if $$A=A^\dagger\quad \Rightarrow \quad \det(A) = \det(A)^*.$$

(Alternatively, the determinant is the product of the matrix's eigenvalues, and as mentioned before, the eigenvalues of a Hermitian matrix are real.)

Hermitian form

References

^ Frankel, Theodore (2004). The geometry of physics: an introduction. Cambridge University Press. p. 652. ISBN 0-521-53927-7.
^ Physics 125 Course Notes at California Institute of Technology