In mathematics, a symmetric matrix ${\displaystyle \ M\ }$ with real entries is positive-definite if the real number ${\displaystyle \ \mathbf {x} ^{\top }M\mathbf {x} \ }$ is positive for every nonzero real column vector ${\displaystyle \ \mathbf {x} \ ,}$ where ${\displaystyle \ \mathbf {x} ^{\top }\ }$ is the row vector transpose of ${\displaystyle \ \mathbf {x} ~.}$^[1] More generally, a Hermitian matrix (that is, a complex matrix equal to its conjugate transpose) is positive-definite if the real number ${\displaystyle \ \mathbf {z} ^{*}M\mathbf {z} \ }$ is positive for every nonzero complex column vector ${\displaystyle \ \mathbf {z} \ ,}$ where ${\displaystyle \ \mathbf {z} ^{*}\ }$ denotes the conjugate transpose of ${\displaystyle \ \mathbf {z} ~.}$

Positive semi-definite matrices are defined similarly, except that the scalars ${\displaystyle \ \mathbf {x} ^{\top }M\mathbf {x} \ }$ and ${\displaystyle \ \mathbf {z} ^{*}M\mathbf {z} \ }$ are required to be positive or zero (that is, nonnegative). Negative-definite and negative semi-definite matrices are defined analogously. A matrix that is not positive semi-definite and not negative semi-definite is sometimes called indefinite.

It follows from the above definitions that a matrix is positive-definite if and only if it is the matrix of a positive-definite quadratic form or Hermitian form. In other words, a matrix is positive-definite if and only if it defines an inner product.

Positive-definite and positive-semidefinite matrices can be characterized in many ways, which may explain the importance of the concept in various parts of mathematics. A matrix M is positive-definite if and only if it satisfies any of the following equivalent conditions.

• ${\displaystyle \ M\ }$ is congruent with a diagonal matrix with positive real entries.
• ${\displaystyle \ M\ }$ is symmetric or Hermitian, and all its eigenvalues are real and positive.
• ${\displaystyle \ M\ }$ is symmetric or Hermitian, and all its leading principal minors are positive.
• There exists an invertible matrix ${\displaystyle \ B\ }$ with conjugate transpose ${\displaystyle \ B^{*}\ }$ such that ${\displaystyle \ M=B^{*}B~.}$

A matrix is positive semi-definite if it satisfies similar equivalent conditions where "positive" is replaced by "nonnegative", "invertible matrix" is replaced by "matrix", and the word "leading" is removed.

Positive-definite and positive-semidefinite real matrices are at the basis of convex optimization, since, given a function of several real variables that is twice differentiable, then if its Hessian matrix (matrix of its second partial derivatives) is positive-definite at a point ${\displaystyle \ p\ ,}$ then the function is convex near p, and, conversely, if the function is convex near ${\displaystyle \ p\ ,}$ then the Hessian matrix is positive-semidefinite at ${\displaystyle \ p~.}$

The set of positive definite matrices is an open convex cone, while the set of positive semi-definite matrices is a closed convex cone.^[2]

Some authors use more general definitions of definiteness, including some non-symmetric real matrices, or non-Hermitian complex ones.

In the following definitions, ${\displaystyle \ \mathbf {x} ^{\top }\ }$ is the transpose of ${\displaystyle \ \mathbf {x} \ ,}$ ${\displaystyle \ \mathbf {z} ^{*}\ }$ is the conjugate transpose of ${\displaystyle \ \mathbf {z} \ ,}$ and ${\displaystyle \ \mathbf {0} \ }$ denotes the n dimensional zero-vector.

An ${\displaystyle n\times n}$ symmetric real matrix ${\displaystyle \ M\ }$ is said to be positive-definite if ${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} >0\ }$ for all non-zero ${\displaystyle \ \mathbf {x} \ }$ in ${\displaystyle \ \mathbb {R} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ symmetric real matrix ${\displaystyle \ M\ }$ is said to be positive-semidefinite or non-negative-definite if ${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} \geq 0\ }$ for all ${\displaystyle \ \mathbf {x} \ }$ in ${\displaystyle \ \mathbb {R} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ symmetric real matrix ${\displaystyle \ M\ }$ is said to be negative-definite if ${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} <0\ }$ for all non-zero ${\displaystyle \ \mathbf {x} \ }$ in ${\displaystyle \ \mathbb {R} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ symmetric real matrix ${\displaystyle \ M\ }$ is said to be negative-semidefinite or non-positive-definite if ${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} \leq 0\ }$ for all ${\displaystyle \ \mathbf {x} \ }$ in ${\displaystyle \ \mathbb {R} ^{n}~.}$ Formally,

An ${\displaystyle n\times n}$ symmetric real matrix which is neither positive semidefinite nor negative semidefinite is called indefinite.

The following definitions all involve the term ${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} ~.}$ Notice that this is always a real number for any Hermitian square matrix ${\displaystyle \ M~.}$

An ${\displaystyle \ n\times n\ }$ Hermitian complex matrix ${\displaystyle \ M\ }$ is said to be positive-definite if ${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} >0\ }$ for all non-zero ${\displaystyle \ \mathbf {z} \ }$ in ${\displaystyle \ \mathbb {C} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ Hermitian complex matrix ${\displaystyle \ M\ }$ is said to be positive semi-definite or non-negative-definite if ${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} \geq 0\ }$ for all ${\displaystyle \ \mathbf {z} \ }$ in ${\displaystyle \ \mathbb {C} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ Hermitian complex matrix ${\displaystyle \ M\ }$ is said to be negative-definite if ${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} <0\ }$ for all non-zero ${\displaystyle \ \mathbf {z} \ }$ in ${\displaystyle \ \mathbb {C} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ Hermitian complex matrix ${\displaystyle \ M\ }$ is said to be negative semi-definite or non-positive-definite if ${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} \leq 0\ }$ for all ${\displaystyle \ \mathbf {z} \ }$ in ${\displaystyle \ \mathbb {C} ^{n}~.}$ Formally,

An ${\displaystyle \ n\times n\ }$ Hermitian complex matrix which is neither positive semidefinite nor negative semidefinite is called indefinite.

Since every real matrix is also a complex matrix, the definitions of "definiteness" for the two classes must agree.

For complex matrices, the most common definition says that ${\displaystyle \ M\ }$ is positive-definite if and only if ${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} \ }$ is real and positive for every non-zero complex column vectors ${\displaystyle \mathbf {z} ~.}$ This condition implies that ${\displaystyle M}$ is Hermitian (i.e. its transpose is equal to its conjugate), since ${\displaystyle \mathbf {z} ^{*}M\ \mathbf {z} }$ being real, it equals its conjugate transpose ${\displaystyle \ \mathbf {z} ^{*}\ M^{*}\ \mathbf {z} \ }$ for every ${\displaystyle \ \mathbf {z} \ ,}$ which implies ${\displaystyle \ M=M^{*}~.}$

By this definition, a positive-definite real matrix ${\displaystyle \ M\ }$ is Hermitian, hence symmetric; and ${\displaystyle \ \mathbf {z} ^{\top }M\ \mathbf {z} \ }$ is positive for all non-zero real column vectors ${\displaystyle \ \mathbf {z} ~.}$ However the last condition alone is not sufficient for ${\displaystyle \ M\ }$ to be positive-definite. For example, if $${\displaystyle \ M={\begin{bmatrix}~1~&~1~\\-1~&~1~\end{bmatrix}},}$$

then for any real vector ${\displaystyle \ \mathbf {z} \ }$ with entries ${\displaystyle \ a\ }$ and ${\displaystyle \ b\ }$ we have ${\displaystyle \ \mathbf {z} ^{\top }M\ \mathbf {z} =\left(a+b\right)a+\left(-a+b\right)b=a^{2}+b^{2}\ ,}$ which is always positive if ${\displaystyle \ \mathbf {z} \ }$ is not zero. However, if ${\displaystyle \ \mathbf {z} \ }$ is the complex vector with entries 1 and ${\displaystyle \ i\ ,}$ one gets

$${\displaystyle \mathbf {z} ^{*}M\ \mathbf {z} ={\begin{bmatrix}~1~&-i~\end{bmatrix}}\ M\ {\begin{bmatrix}~1~\\~i~\end{bmatrix}}={\begin{bmatrix}~1+i~&~1-i~\end{bmatrix}}\ {\begin{bmatrix}~1~\\~i~\end{bmatrix}}=2+2i~.}$$

which is not real. Therefore, ${\displaystyle \ M\ }$ is not positive-definite.

On the other hand, for a symmetric real matrix ${\displaystyle \ M\ ,}$ the condition "${\displaystyle \ \mathbf {z} ^{\top }M\ \mathbf {z} >0\ }$ for all nonzero real vectors ${\displaystyle \ \mathbf {z} \ }$ does imply that ${\displaystyle \ M\ }$ is positive-definite in the complex sense.

If a Hermitian matrix ${\displaystyle \ M\ }$ is positive semi-definite, one sometimes writes ${\displaystyle \ M\succeq 0\ }$ and if ${\displaystyle \ M\ }$ is positive-definite one writes ${\displaystyle \ M\succ 0~.}$ To denote that ${\displaystyle \ M\ }$ is negative semi-definite one writes ${\displaystyle \ M\preceq 0\ }$ and to denote that ${\displaystyle \ M\ }$ is negative-definite one writes ${\displaystyle \ M\prec 0~.}$

The notion comes from functional analysis where positive semidefinite matrices define positive operators. If two matrices ${\displaystyle \ A\ }$ and ${\displaystyle \ B\ }$ satisfy ${\displaystyle \ B-A\succeq 0\ ,}$ we can define a non-strict partial order ${\displaystyle \ B\succeq A\ }$ that is reflexive, antisymmetric, and transitive; It is not a total order, however, as ${\displaystyle \ B-A\ ,}$ in general, may be indefinite.

A common alternative notation is ${\displaystyle \ M\geq 0\ ,}$ ${\displaystyle \ M>0\ ,}$ ${\displaystyle \ M\leq 0\ ,}$ and ${\displaystyle \ M<0\ }$ for positive semi-definite and positive-definite, negative semi-definite and negative-definite matrices, respectively. This may be confusing, as sometimes nonnegative matrices (respectively, nonpositive matrices) are also denoted in this way.

Let ${\displaystyle M}$ be an ${\displaystyle n\times n}$ Hermitian matrix (this includes real symmetric matrices). All eigenvalues of ${\displaystyle M}$ are real, and their sign characterize its definiteness:

• ${\displaystyle M}$ is positive definite if and only if all of its eigenvalues are positive.
• ${\displaystyle M}$ is positive semi-definite if and only if all of its eigenvalues are non-negative.
• ${\displaystyle M}$ is negative definite if and only if all of its eigenvalues are negative
• ${\displaystyle M}$ is negative semi-definite if and only if all of its eigenvalues are non-positive.
• ${\displaystyle M}$ is indefinite if and only if it has both positive and negative eigenvalues.

Let ${\displaystyle \ PDP^{-1}\ }$ be an eigendecomposition of ${\displaystyle \ M\ ,}$ where ${\displaystyle \ P\ }$ is a unitary complex matrix whose columns comprise an orthonormal basis of eigenvectors of ${\displaystyle \ M\ ,}$ and ${\displaystyle \ D\ }$ is a real diagonal matrix whose main diagonal contains the corresponding eigenvalues. The matrix ${\displaystyle \ M\ }$ may be regarded as a diagonal matrix ${\displaystyle \ D\ }$ that has been re-expressed in coordinates of the (eigenvectors) basis ${\displaystyle \ P~.}$ Put differently, applying ${\displaystyle M}$ to some vector ${\displaystyle \ \mathbf {z} \ ,}$ giving ${\displaystyle \ M\mathbf {z} \ ,}$ is the same as changing the basis to the eigenvector coordinate system using ${\displaystyle \ P^{-1}\ ,}$ giving ${\displaystyle \ P^{-1}\mathbf {z} \ ,}$ applying the stretching transformation ${\displaystyle \ D\ }$ to the result, giving ${\displaystyle \ DP^{-1}\mathbf {z} \ ,}$ and then changing the basis back using ${\displaystyle \ P\ ,}$ giving ${\displaystyle \ PDP^{-1}\mathbf {z} ~.}$

With this in mind, the one-to-one change of variable ${\displaystyle \ \mathbf {y} =P\mathbf {z} \ }$ shows that ${\displaystyle \ \mathbf {z} ^{*}M\mathbf {z} \ }$ is real and positive for any complex vector ${\displaystyle \ \mathbf {z} \ }$ if and only if ${\displaystyle \ \mathbf {y} ^{*}D\mathbf {y} \ }$ is real and positive for any ${\displaystyle \ y\ ;}$ in other words, if ${\displaystyle \ D\ }$ is positive definite. For a diagonal matrix, this is true only if each element of the main diagonal – that is, every eigenvalue of ${\displaystyle \ M\ }$ – is positive. Since the spectral theorem guarantees all eigenvalues of a Hermitian matrix to be real, the positivity of eigenvalues can be checked using Descartes' rule of alternating signs when the characteristic polynomial of a real, symmetric matrix ${\displaystyle \ M\ }$ is available.

Let ${\displaystyle \ M\ }$ be an ${\displaystyle \ n\times n\ }$ Hermitian matrix. ${\displaystyle \ M\ }$ is positive semidefinite if and only if it can be decomposed as a product $${\displaystyle \ M=B^{*}B\ }$$ of a matrix ${\displaystyle \ B\ }$ with its conjugate transpose.

When ${\displaystyle \ M\ }$ is real, ${\displaystyle \ B\ }$ can be real as well and the decomposition can be written as $${\displaystyle \ M=B^{\top }B~.}$$

${\displaystyle M}$ is positive definite if and only if such a decomposition exists with ${\displaystyle \ B\ }$ invertible. More generally, ${\displaystyle \ M\ }$ is positive semidefinite with rank ${\displaystyle \ k\ }$ if and only if a decomposition exists with a ${\displaystyle \ k\times n\ }$ matrix ${\displaystyle \ B\ }$ of full row rank (i.e. of rank ${\displaystyle \ k\ }$). Moreover, for any decomposition ${\displaystyle \ M=B^{*}B\ ,}$ ${\displaystyle \ \operatorname {rank} (M)=\operatorname {rank} (B)~.}$^[3]

The columns ${\displaystyle \ b_{1},\dots ,b_{n}\ }$ of ${\displaystyle \ B\ }$ can be seen as vectors in the complex or real vector space ${\displaystyle \ \mathbb {R} ^{k}\ ,}$ respectively. Then the entries of ${\displaystyle M}$ are inner products (that is dot products, in the real case) of these vectors $${\displaystyle \ M_{ij}=\langle b_{i},b_{j}\rangle ~.}$$ In other words, a Hermitian matrix ${\displaystyle \ M\ }$ is positive semidefinite if and only if it is the Gram matrix of some vectors ${\displaystyle \ b_{1},\dots ,b_{n}~.}$ 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 ${\displaystyle \ b_{1},\dots ,b_{n}\ }$ equals the dimension of the space spanned by these vectors.^[4]

The decomposition is not unique: if ${\displaystyle \ M=B^{*}B\ }$ for some ${\displaystyle \ k\times n\ }$ matrix ${\displaystyle \ B\ }$ and if ${\displaystyle \ Q\ }$ is any unitary ${\displaystyle k\times k}$ matrix (meaning ${\displaystyle \ Q^{*}Q=QQ^{*}=I\ }$), then ${\displaystyle \ M=B^{*}B=B^{*}Q^{*}QB=A^{*}A\ }$ for ${\displaystyle \ A=QB~.}$

However, this is the only way in which two decompositions can differ: The decomposition is unique up to unitary transformations. More formally, if ${\displaystyle \ A\ }$ is a ${\displaystyle \ k\times n\ }$ matrix and ${\displaystyle \ B\ }$ is a ${\displaystyle \ \ell \times n\ }$ matrix such that ${\displaystyle \ A^{*}A=B^{*}B\ ,}$ then there is a ${\displaystyle \ \ell \times k\ }$ matrix ${\displaystyle Q}$ with orthonormal columns (meaning ${\displaystyle \ Q^{*}Q=I_{k\times k}\ }$) such that ${\displaystyle \ B=QA~.}$^[5] When ${\displaystyle \ell =k}$ this means ${\displaystyle Q}$ is unitary.

This statement has an intuitive geometric interpretation in the real case: let the columns of ${\displaystyle A}$ and ${\displaystyle B}$ be the vectors ${\displaystyle a_{1},\dots ,a_{n}}$ and ${\displaystyle \ b_{1},\dots ,b_{n}\ }$ in ${\displaystyle \ \mathbb {R} ^{k}~.}$ A real unitary matrix is an orthogonal matrix, which describes a rigid transformation (an isometry of Euclidean space ${\displaystyle \mathbb {R} ^{k}}$) preserving the 0 point (i.e. rotations and reflections, without translations). Therefore, the dot products ${\displaystyle a_{i}\cdot a_{j}}$ and ${\displaystyle b_{i}\cdot b_{j}}$ are equal if and only if some rigid transformation of ${\displaystyle \mathbb {R} ^{k}}$ transforms the vectors ${\displaystyle a_{1},\dots ,a_{n}}$ to ${\displaystyle b_{1},\dots ,b_{n}}$ (and 0 to 0).

A Hermitian matrix ${\displaystyle \ M\ }$ is positive semidefinite if and only if there is a positive semidefinite matrix ${\displaystyle \ B\ }$ (in particular ${\displaystyle \ B\ }$ is Hermitian, so ${\displaystyle \ B^{*}=B\ }$) satisfying ${\displaystyle \ M=BB~.}$ This matrix ${\displaystyle \ B\ }$ is unique,^[6] is called the non-negative square root of ${\displaystyle \ M\ ,}$ and is denoted with ${\displaystyle \ B=M^{\frac {1}{2}}~.}$ When ${\displaystyle M}$ is positive definite, so is ${\displaystyle \ M^{\frac {1}{2}}\ ,}$ hence it is also called the positive square root of ${\displaystyle \ M~.}$

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

If ${\displaystyle \ M\succ N\succ 0\ }$ then ${\displaystyle \ M^{\frac {1}{2}}\succ N^{\frac {1}{2}}\succ 0~.}$

A Hermitian positive semidefinite matrix ${\displaystyle \ M\ }$ can be written as ${\displaystyle \ M=LL^{*}\ ,}$ where ${\displaystyle \ L\ }$ is lower triangular with non-negative diagonal (equivalently ${\displaystyle M=B^{*}B}$ where ${\displaystyle \ B=L^{*}\ }$ is upper triangular); this is the Cholesky decomposition. If ${\displaystyle \ M\ }$ is positive definite, then the diagonal of ${\displaystyle \ L\ }$ is positive and the Cholesky decomposition is unique. Conversely if ${\displaystyle \ L\ }$ is lower triangular with nonnegative diagonal then ${\displaystyle \ LL^{*}\ }$ is positive semidefinite. The Cholesky decomposition is especially useful for efficient numerical calculations. A closely related decomposition is the LDL decomposition, ${\displaystyle \ M=LDL^{*}\ ,}$ where ${\displaystyle D}$ is diagonal and ${\displaystyle \ L\ }$ is lower unitriangular.

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

Let ${\displaystyle \ M\ }$ be an ${\displaystyle \ n\times n\ }$ real symmetric matrix, and let ${\displaystyle \ B_{1}(M)\equiv \{\mathbf {x} \in \mathbb {R} ^{n}:\mathbf {x} ^{\top }M\ \mathbf {x} \leq 1\}\ }$ be the "unit ball" defined by ${\displaystyle \ M~.}$ Then we have the following