Main diagonal values are real The entries on the
main diagonal (top left to bottom right) of any Hermitian matrix are
real. {{math proof|1= By definition of the Hermitian matrix H_{ij} = \overline{H}_{ji} so for the above follows, as a number can equal its complex conjugate only if the imaginary parts are zero. }} Only the
main diagonal entries are necessarily real; Hermitian matrices can have arbitrary complex-valued entries in their
off-diagonal elements, as long as diagonally-opposite entries are complex conjugates.
Symmetric A matrix that has only real entries is
symmetric if and only if it is a Hermitian matrix. A real and symmetric matrix is simply a special case of a Hermitian matrix. {{math proof|1= H_{ij} = \overline{H}_{ji} by definition. Thus H_{ij} = H_{ji} (matrix symmetry) if and only if H_{ij} = \overline{H}_{ij} (H_{ij} is real). }} So, if a real anti-symmetric matrix is multiplied by a real multiple of the imaginary unit i, then it becomes Hermitian.
Normal Every Hermitian matrix is a
normal matrix. That is to say, AA^\mathsf{H} = A^\mathsf{H}A. {{math proof|1=A = A^\mathsf{H}, so AA^\mathsf{H} = AA = A^\mathsf{H}A.}}
Diagonalizable The finite-dimensional
spectral theorem 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 with dimension are real, and that has linearly independent
eigenvectors. Moreover, a Hermitian matrix has
orthogonal eigenvectors for distinct eigenvalues. Even if there are degenerate eigenvalues, it is always possible to find an
orthogonal basis of consisting of eigenvectors of .
Sum of Hermitian matrices The sum of any two Hermitian matrices is Hermitian. {{math proof|1= (A + B)_{ij} = A_{ij} + B_{ij} = \overline{A}_{ji} + \overline{B}_{ji} = \overline{(A + B)}_{ji}, as claimed.}}
Inverse is Hermitian The
inverse of an invertible Hermitian matrix is Hermitian as well. {{math proof|1= If A^{-1}A = I, then I= I^\mathsf{H} = \left(A^{-1}A\right)^\mathsf{H} = A^\mathsf{H}\left(A^{-1}\right)^\mathsf{H} = A \left(A^{-1}\right)^\mathsf{H}, so A^{-1}=\left(A^{-1}\right)^\mathsf{H} as claimed.}}
Associative product of Hermitian matrices The
product of two Hermitian matrices and is Hermitian if and only if . {{math proof|1= (AB)^\mathsf{H} = \overline{(AB)^\mathsf{T}} = \overline{B^\mathsf{T} A^\mathsf{T}} = \overline{B^\mathsf{T}} \ \overline{A^\mathsf{T}} = B^\mathsf{H} A^\mathsf{H} = BA. Thus (AB)^\mathsf{H} = AB
if and only if AB = BA. Thus is Hermitian if is Hermitian and is an integer. }}
ABA Hermitian If
A and
B are Hermitian, then
ABA is also Hermitian. {{math proof|1= (ABA)^\mathsf{H} = (A(BA))^\mathsf{H} = (BA)^\mathsf{H}A^\mathsf{H} = A^\mathsf{H}B^\mathsf{H}A^\mathsf{H} = ABA }}
is real for complex For an arbitrary complex valued vector the product \mathbf{v}^\mathsf{H} A \mathbf{v} is real because of \mathbf{v}^\mathsf{H} A \mathbf{v} = \left(\mathbf{v}^\mathsf{H} A \mathbf{v}\right)^\mathsf{H} . This is especially important in quantum physics where Hermitian matrices are operators that measure properties of a system, e.g. total
spin, which have to be real.
Complex Hermitian forms vector space over The Hermitian complex -by- matrices do not form a
vector space over the
complex numbers, , since the identity matrix is Hermitian, but is not. However the complex Hermitian matrices
do form a vector space over the
real numbers . In the -
dimensional vector space of complex matrices over , the complex Hermitian matrices form a subspace of dimension . If denotes the -by- matrix with a in the position and zeros elsewhere, a basis (orthonormal with respect to the Frobenius inner product) can be described as follows: E_{jj} \text{ for } 1 \leq j \leq n \quad (n \text{ matrices}) together with the set of matrices of the form \frac{1}{\sqrt{2}}\left(E_{jk} + E_{kj}\right) \text{ for } 1 \leq j and the matrices \frac{i}{\sqrt{2}}\left(E_{jk} - E_{kj}\right) \text{ for } 1 \leq j where i denotes the
imaginary unit, i = \sqrt{-1}~. An example is that the four
Pauli matrices form a complete basis for the vector space of all complex 2-by-2 Hermitian matrices over .
Eigendecomposition If orthonormal eigenvectors \mathbf{u}_1, \dots, \mathbf{u}_n of a Hermitian matrix are chosen and written as the columns of the matrix , then one
eigendecomposition of is A = U \Lambda U^\mathsf{H} where U U^\mathsf{H} = I = U^\mathsf{H} U and therefore A = \sum_j \lambda_j \mathbf{u}_j \mathbf{u}_j ^\mathsf{H}, where \lambda_j are the eigenvalues on the diagonal of the diagonal matrix \Lambda.
Singular values The singular values of A are the absolute values of its eigenvalues: Since A has an eigendecomposition A=U\Lambda U^H, where U is a
unitary matrix (its columns are orthonormal vectors;
see above), a
singular value decomposition of A is A=U|\Lambda|\text{sgn}(\Lambda)U^H, where |\Lambda| and \text{sgn}(\Lambda) are diagonal matrices containing the absolute values |\lambda| and signs \text{sgn}(\lambda) of A's eigenvalues, respectively. \sgn(\Lambda)U^H is unitary, since the columns of U^H are only getting multiplied by \pm 1. |\Lambda| contains the singular values of A, namely, the absolute values of its eigenvalues.
Real determinant The determinant of a Hermitian matrix is real: {{math proof|1= \det(A) = \det\left(A^\mathsf{T}\right)\quad \Rightarrow \quad \det\left(A^\mathsf{H}\right) = \overline{\det(A)} Therefore if A = A^\mathsf{H}\quad \Rightarrow \quad \det(A) = \overline{\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.) ==Decomposition into Hermitian and skew-Hermitian matrices==