Editing Hermitian matrix (section)

==Properties==

===Main diagonal values are real===

The entries on the [[main diagonal]] (top left to bottom right) of any Hermitian matrix are [[real number|real]].

{{math proof|1= By definition of the Hermitian matrix
<math display=block>H_{ij} = \overline{H}_{ji} </math>
so for {{math|1=''i'' = ''j''}} 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 element]]s, as long as diagonally-opposite entries are complex conjugates.

===Symmetric===

A matrix that has only real entries is [[symmetric matrix|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= <math>H_{ij} = \overline{H}_{ji}</math> by definition. Thus  <math>H_{ij} = H_{ji}</math> (matrix symmetry) if and only if <math>H_{ij} = \overline{H}_{ij}</math> (<math>H_{ij}</math> is real).
}}

So, if a real anti-symmetric matrix is multiplied by a real multiple of the imaginary unit <math>i,</math> then it becomes Hermitian.

===Normal===

Every Hermitian matrix is a [[normal matrix]]. That is to say, <math>AA^\mathsf{H} = A^\mathsf{H}A.</math>

{{math proof|1=<math>A = A^\mathsf{H},</math> so <math>AA^\mathsf{H} = AA = A^\mathsf{H}A.</math>}}

===Diagonalizable===

The finite-dimensional [[spectral theorem]] says that any Hermitian matrix can be [[diagonalizable matrix|diagonalized]] by a [[unitary matrix]], and that the resulting diagonal matrix has only real entries. This implies that all [[eigenvectors|eigenvalue]]s of a Hermitian matrix {{mvar|A}} with dimension {{mvar|n}} are real, and that {{mvar|A}} has {{mvar|n}} linearly independent [[eigenvector]]s. 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 {{math|'''C'''<sup>''n''</sup>}} consisting of {{mvar|n}} eigenvectors of {{mvar|A}}.

===Sum of Hermitian matrices===

The sum of any two Hermitian matrices is Hermitian.

{{math proof|1= <math display="block">(A + B)_{ij} = A_{ij} + B_{ij} = \overline{A}_{ji} + \overline{B}_{ji} = \overline{(A + B)}_{ji},</math> as claimed.}}

===Inverse is Hermitian===

The [[inverse matrix|inverse]] of an invertible Hermitian matrix is Hermitian as well.

{{math proof|1= If <math>A^{-1}A = I,</math> then <math>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},</math> so <math>A^{-1}=\left(A^{-1}\right)^\mathsf{H}</math> as claimed.}}

===Associative product of Hermitian matrices===

The [[matrix multiplication|product]] of two Hermitian matrices {{mvar|A}} and {{mvar|B}} is Hermitian if and only if {{math|1=''AB'' = ''BA''}}.

{{math proof|1= <math display="block">(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.</math> Thus <math>(AB)^\mathsf{H} = AB</math> [[if and only if]] <math>AB = BA.</math>

Thus {{math|''A''<sup>''n''</sup>}} is Hermitian if {{mvar|A}} is Hermitian and {{mvar|n}} is an integer.
}}

===''ABA'' Hermitian===

If ''A'' and ''B'' are Hermitian, then ''ABA'' is also Hermitian.
{{math proof|1= <math display="block">(ABA)^\mathsf{H} = (A(BA))^\mathsf{H} = (BA)^\mathsf{H}A^\mathsf{H} = A^\mathsf{H}B^\mathsf{H}A^\mathsf{H} = ABA </math>}}

==={{math|v<sup>H</sup>''A''v}} is real for complex {{math|v}}===

For an arbitrary complex valued vector {{Math|'''v'''}} the product <math> \mathbf{v}^\mathsf{H} A \mathbf{v} </math> is real because of <math> \mathbf{v}^\mathsf{H} A \mathbf{v} = \left(\mathbf{v}^\mathsf{H} A \mathbf{v}\right)^\mathsf{H} .</math> This is especially important in quantum physics where Hermitian matrices are operators that measure properties of a system, e.g. total [[Spin (physics)|spin]], which have to be real.

===Complex Hermitian forms vector space over {{math|ℝ}}===

The Hermitian complex {{mvar|n}}-by-{{mvar|n}} matrices do not form a [[vector space]] over the [[complex number]]s, {{math|'''ℂ'''}}, since the identity matrix {{math|''I''<sub>''n''</sub>}} is Hermitian, but {{math|''i'' ''I''<sub>''n''</sub>}} is not. However the complex Hermitian matrices ''do'' form a vector space over the [[real numbers]] {{math|'''ℝ'''}}. In the {{math|2''n''<sup>2</sup>}}-[[dimension of a vector space|dimensional]] vector space of complex {{math|''n'' × ''n''}} matrices over {{math|'''ℝ'''}}, the complex Hermitian matrices form a subspace of dimension {{math|''n''<sup>2</sup>}}. If {{math|''E''<sub>''jk''</sub>}} denotes the {{mvar|n}}-by-{{mvar|n}} matrix with a {{math|1}} in the {{math|''j'',''k''}} position and zeros elsewhere, a basis (orthonormal with respect to the Frobenius inner product) can be described as follows:
<math display=block>E_{jj} \text{ for } 1 \leq j \leq n \quad (n \text{ matrices}) </math>

together with the set of matrices of the form
<math display=block>\frac{1}{\sqrt{2}}\left(E_{jk} + E_{kj}\right) \text{ for } 1 \leq j < k \leq n \quad \left( \frac{n^2-n} 2 \text{ matrices} \right) </math>

and the matrices
<math display=block>\frac{i}{\sqrt{2}}\left(E_{jk} - E_{kj}\right) \text{ for } 1 \leq j < k \leq n \quad \left( \frac{n^2-n} 2 \text{ matrices} \right) </math>

where <math>i</math> denotes the [[imaginary unit]], <math>i = \sqrt{-1}~.</math>

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 {{math|'''ℝ'''}}.

===Eigendecomposition===

If {{mvar|n}} orthonormal eigenvectors <math>\mathbf{u}_1, \dots, \mathbf{u}_n</math> of a Hermitian matrix are chosen and written as the columns of the matrix {{mvar|U}}, then one [[Eigendecomposition of a matrix|eigendecomposition]] of {{mvar|A}} is <math> A = U \Lambda U^\mathsf{H}</math> where <math>U U^\mathsf{H} = I = U^\mathsf{H} U</math> and therefore 
<math display=block>A = \sum_j \lambda_j \mathbf{u}_j \mathbf{u}_j ^\mathsf{H},</math>
where <math>\lambda_j</math> are the eigenvalues on the diagonal of the diagonal matrix <math>\Lambda.</math>

=== Singular values ===
The singular values of <math>A</math> are the absolute values of its eigenvalues:

Since <math>A</math> has an eigendecomposition <math>A=U\Lambda U^H</math>, where <math>U</math> is a [[unitary matrix]] (its columns are orthonormal vectors; [[Hermitian matrix#Eigendecomposition|see above]]), a [[singular value decomposition]] of <math>A</math> is <math>A=U|\Lambda|\text{sgn}(\Lambda)U^H</math>, where <math>|\Lambda|</math> and <math>\text{sgn}(\Lambda)</math> are diagonal matrices containing the absolute values <math>|\lambda|</math> and signs <math>\text{sgn}(\lambda)</math> of <math>A</math>'s eigenvalues, respectively. <math>\sgn(\Lambda)U^H</math> is unitary, since the columns of <math>U^H</math> are only getting multiplied by <math>\pm 1</math>. <math>|\Lambda|</math> contains the singular values of <math>A</math>, namely, the absolute values of its eigenvalues.<ref>{{Cite book |last1=Trefethan |first1=Lloyd N. |url=http://worldcat.org/oclc/1348374386 |title=Numerical linear algebra |last2=Bau, III |first2=David |publisher=[[SIAM]] |year=1997 |isbn=0-89871-361-7 |location=Philadelphia, PA, USA |pages=34 |oclc=1348374386}}</ref>

===Real determinant===

The determinant of a Hermitian matrix is real:

{{math proof|1= <math> \det(A) = \det\left(A^\mathsf{T}\right)\quad \Rightarrow \quad \det\left(A^\mathsf{H}\right) = \overline{\det(A)} </math>
Therefore if <math>A = A^\mathsf{H}\quad \Rightarrow \quad \det(A) = \overline{\det(A)} .</math>
}}
(Alternatively, the determinant is the product of the matrix's eigenvalues, and as mentioned before, the eigenvalues of a Hermitian matrix are real.)