Editing Definite matrix (section)

== Decomposition ==
{{See also|Gram matrix}}
Let <math>M</math> be an <math>n \times n</math> [[Hermitian matrix]].
<math>M</math> is positive semidefinite if and only if it can be decomposed as a product
<math display="block">M = B^* B</math>
of a matrix <math>B</math> with its [[conjugate transpose]].

When <math>M</math> is real, <math>B</math> can be real as well and the decomposition can be written as <math display="block">M = B^\mathsf{T} B.</math>

<math>M</math> is positive definite if and only if such a decomposition exists with <math>B</math> [[Invertible matrix|invertible]].
More generally, <math>M</math> is positive semidefinite with rank <math>k</math> if and only if a decomposition exists with a <math>k \times n</math> matrix <math>B</math> of full row rank (i.e. of rank <math>k</math>).
Moreover, for any decomposition <math>M = B^* B,</math> <math>\operatorname{rank}(M) = \operatorname{rank}(B).</math><ref>{{harvtxt|Horn|Johnson|2013}}, p. 440, Theorem 7.2.7</ref>

{{math proof | proof =
If <math>M = B^* B,</math> then <math>x^* M x = (x^* B^*) (B x) = \|B x \|^2 \geq 0,</math> so <math>M</math> is positive semidefinite.
If moreover <math>B</math> is invertible then the inequality is strict for <math>x \neq 0,</math> so <math>M</math> is positive definite.
If <math>B</math> is <math>k \times n</math> of rank <math>k,</math> then <math>\operatorname{rank}(M) = \operatorname{rank}(B^*) = k.</math>

In the other direction, suppose <math>M</math> is positive semidefinite.
Since <math>M</math> is Hermitian, it has an [[Eigendecomposition of a matrix#Decomposition for special matrices|eigendecomposition]] <math>M = Q^{-1} D Q</math> where <math>Q</math> is [[unitary matrix|unitary]] and <math>D</math> is a diagonal matrix whose entries are the eigenvalues of <math>M</math>
Since <math>M</math> is positive semidefinite, the eigenvalues are non-negative real numbers, so one can define <math>D^{\frac{1}{2}}</math> as the diagonal matrix whose entries are non-negative square roots of eigenvalues.
Then <math>M = Q^{-1} D Q = Q^* D Q = Q^* D^{\frac{1}{2}} D^{\frac{1}{2}} Q = Q^* D^{\frac{1}{2}*} D^{\frac{1}{2}} Q = B^* B</math> for <math>B = D^{\frac{1}{2}} Q.</math>
If moreover <math>M</math> is positive definite, then the eigenvalues are (strictly) positive, so <math>D^{\frac{1}{2}}</math> is invertible, and hence <math>B = D^{\frac{1}{2}} Q</math> is invertible as well.
If <math>M</math> has rank <math>k,</math> then it has exactly <math>k</math> positive eigenvalues and the others are zero, hence in <math>B = D^{\frac{1}{2}} Q</math> all but <math>k</math> rows are all zeroed.
Cutting the zero rows gives a <math>k \times n</math> matrix <math>B'</math> such that <math>B'^* B' = B^* B = M.</math>
}}

The columns <math>b_1, \dots, b_n</math> of <math>B</math> can be seen as vectors in the [[complex vector space|complex]] or [[real vector space]] <math>\mathbb{R}^k,</math> respectively.
Then the entries of <math>M</math> are [[inner product]]s (that is [[dot product]]s, in the real case) of these vectors
<math display="block">M_{ij} = \langle b_i, b_j\rangle.</math>
In other words, a Hermitian matrix <math>M</math> is positive semidefinite if and only if it is the [[Gram matrix]] of some vectors <math>b_1, \dots, b_n.</math>
It is positive definite if and only if it is the Gram matrix of some [[linearly independent]] vectors.
In general, the rank of the Gram matrix of vectors <math>b_1, \dots, b_n</math> equals the dimension of the space [[Linear span|spanned]] by these vectors.<ref>{{harvtxt|Horn|Johnson|2013}}, p. 441, Theorem 7.2.10</ref>

=== Uniqueness up to unitary transformations ===
The decomposition is not unique: 
if <math>M = B^* B</math> for some <math>k \times n</math> matrix <math>B</math> and if <math>Q</math> is any [[unitary matrix|unitary]] <math>k \times k</math> matrix (meaning <math>Q^* Q = Q Q^* = I</math>),
then <math>M = B^* B = B^* Q^* Q B = A^* A</math> for <math>A = Q B.</math>

However, this is the only way in which two decompositions can differ: The decomposition is unique up to [[unitary transformation]]s.
More formally, if <math>A</math> is a <math>k \times n</math> matrix and <math>B</math> is a <math>\ell \times n</math> matrix such that <math>A^* A = B^* B,</math>
then there is a <math>\ell \times k</math> matrix <math>Q</math> with orthonormal columns (meaning <math>Q^* Q = I_{k \times k}</math>) such that <math>B = Q A.</math><ref>{{harvtxt|Horn|Johnson|2013}}, p. 452, Theorem 7.3.11</ref>
When <math>\ell = k</math> this means <math>Q</math> is [[unitary matrix|unitary]].

This statement has an intuitive geometric interpretation in the real case:
let the columns of <math>A</math> and <math>B</math> be the vectors <math>a_1,\dots,a_n</math> and <math>b_1, \dots, b_n</math> in <math>\mathbb{R}^k.</math>
A real unitary matrix is an [[orthogonal matrix]], which describes a [[rigid transformation]] (an isometry of Euclidean space <math>\mathbb{R}^k</math>) preserving the 0 point (i.e. [[Rotation matrix|rotations]] and [[Reflection matrix|reflections]], without translations). 
Therefore, the dot products <math>a_i \cdot a_j</math> and <math>b_i \cdot b_j</math> are equal if and only if some rigid transformation of <math>\mathbb{R}^k</math> transforms the vectors <math>a_1,\dots,a_n</math> to <math>b_1,\dots,b_n</math> (and 0 to 0).

=== Square root ===
{{main|Square root of a matrix}}
A Hermitian matrix <math>M</math> is positive semidefinite if and only if there is a positive semidefinite matrix <math>B</math>  (in particular <math>B</math> is Hermitian, so <math>B^* = B</math>) satisfying <math>M = B B.</math> This matrix <math>B</math> is unique,<ref>{{harvtxt|Horn|Johnson|2013}}, p. 439, Theorem 7.2.6 with <math>k = 2</math></ref> is called the ''non-negative [[square root of a matrix|square root]]'' of <math>M,</math> and is denoted with <math>B = M^\frac{1}{2}.</math>
When <math>M</math> is positive definite, so is <math>M^\frac{1}{2},</math> hence it is also called the ''positive square root'' of <math>M .</math>

The non-negative square root should not be confused with other decompositions <math>M = B^* B.</math>
Some authors use the name ''square root'' and <math>M^\frac{1}{2}</math> for any such decomposition, or specifically for the [[Cholesky decomposition]],
or any decomposition of the form <math>M = B B;</math>
others only use it for the non-negative square root.

If <math>M \succ N \succ 0</math> then <math>M^\frac{1}{2} \succ N^\frac{1}{2} \succ 0.</math>

=== Cholesky decomposition ===
A Hermitian positive semidefinite matrix <math>M</math> can be written as <math>M = L L^*,</math> where <math>L</math> is lower triangular with non-negative diagonal (equivalently <math>M = B^*B</math> where <math>B = L^*</math> is upper triangular); this is the [[Cholesky decomposition]].
If <math>M</math> is positive definite, then the diagonal of <math>L</math> is positive and the Cholesky decomposition is unique. Conversely if <math>L</math> is lower triangular with nonnegative diagonal then <math>L L^*</math> is positive semidefinite. 
The Cholesky decomposition is especially useful for efficient numerical calculations.
A closely related decomposition is the [[Cholesky decomposition#LDL decomposition|LDL decomposition]], <math>M = L D L^*,</math> where <math>D</math> is diagonal and <math>L</math> is [[Triangular matrix#Unitriangular matrix|lower unitriangular]].

=== Williamson theorem ===
Any <math>2n\times 2n </math> positive definite Hermitian real matrix <math>M </math> can be diagonalized via symplectic (real) matrices. More precisely, [[Williamson theorem|Williamson's theorem]] ensures the existence of symplectic <math>S\in\mathbf{Sp}(2n,\mathbb{R}) </math> and diagonal real positive <math>D\in\mathbb{R}^{n\times n} </math> such that <math>SMS^T=D\oplus D </math>.