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Definite matrix

Inmathematics, 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 transposeof${\displaystyle \ \mathbf {x} ~.}$^[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, not negative). 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.

Ramifications[edit]

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 formorHermitian 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.

• ${\displaystyle \ M\ }$iscongruent 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.

Definitions[edit]

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

Definitions for real matrices[edit]

An${\displaystyle n\times n}$ symmetric real matrix ${\displaystyle \ M\ }$ is said to be positive-definiteif${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} >0\ }$ 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-semidefiniteornon-negative-definiteif${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} \geq 0\ }$ 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-definiteif${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} <0\ }$ 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-semidefiniteornon-positive-definiteif${\displaystyle \ \mathbf {x} ^{\top }M\ \mathbf {x} \leq 0\ }$ 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.

Definitions for complex matrices[edit]

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-definiteif${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} >0\ }$ 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-definiteornon-negative-definiteif${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} \geq 0\ }$ 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-definiteif${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} <0\ }$ 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-definiteornon-positive-definiteif${\displaystyle \ \mathbf {z} ^{*}M\ \mathbf {z} \leq 0\ }$ 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.

Consistency between real and complex definitions[edit]

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

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

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.

Notation[edit]

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.

Examples[edit]

${\displaystyle \mathbf {z} }$

${\displaystyle a}$

${\displaystyle b}$

The real symmetric matrix

$${\displaystyle M={\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}}}$$

is positive-definite since for any non-zero column vector z with entries a, b and c, we have

$${\displaystyle {\begin{aligned}\mathbf {z} ^{\top }M\mathbf {z} =\left(\mathbf {z} ^{\top }M\right)\mathbf {z} &={\begin{bmatrix}(2a-b)&(-a+2b-c)&(-b+2c)\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}\\&=(2a-b)a+(-a+2b-c)b+(-b+2c)c\\&=2a^{2}-ba-ab+2b^{2}-cb-bc+2c^{2}\\&=2a^{2}-2ab+2b^{2}-2bc+2c^{2}\\&=a^{2}+a^{2}-2ab+b^{2}+b^{2}-2bc+c^{2}+c^{2}\\&=a^{2}+(a-b)^{2}+(b-c)^{2}+c^{2}\end{aligned}}}$$

This result is a sum of squares, and therefore non-negative; and is zero only if ${\displaystyle \ a=b=c=0\ ,}$ that is, when ${\displaystyle \ \mathbf {z} \ }$ is the zero vector.

For any real invertible matrix ${\displaystyle \ A\ ,}$ the product ${\displaystyle \ A^{\top }A\ }$ is a positive definite matrix (if the means of the columns of A are 0, then this is also called the covariance matrix). A simple proof is that for any non-zero vector ${\displaystyle \ \mathbf {z} \ ,}$ the condition ${\displaystyle \ \mathbf {z} ^{\top }A^{\top }A\mathbf {z} =(A\mathbf {z} )^{\top }(A\mathbf {z} )=\|A\mathbf {z} \|^{2}>0\ ,}$ since the invertibility of matrix ${\displaystyle A}$ means that ${\displaystyle A\mathbf {z} \neq 0~.}$

The example ${\displaystyle M}$ above shows that a matrix in which some elements are negative may still be positive definite. Conversely, a matrix whose entries are all positive is not necessarily positive definite, as for example

$${\displaystyle \ N={\begin{bmatrix}1&2\\2&1\end{bmatrix}}\ ,}$$

for which ${\displaystyle {\begin{bmatrix}-1&1\end{bmatrix}}N{\begin{bmatrix}-1&1\end{bmatrix}}^{\top }=-2<0~.}$

Eigenvalues[edit]

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 eigendecompositionof${\displaystyle \ M\ ,}$ where ${\displaystyle \ P\ }$ is a unitary complex matrix whose columns comprise an orthonormal basisofeigenvectorsof${\displaystyle \ M\ ,}$ 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.

Decomposition[edit]

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

${\displaystyle \ B\ }$

When ${\displaystyle \ M\ }$ is real, ${\displaystyle \ B\ }$ can be real as well and the decomposition can be written as

${\displaystyle M}$ 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]

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

The columns ${\displaystyle \ b_{1},\dots ,b_{n}\ }$of${\displaystyle \ B\ }$ can be seen as vectors in the complexorreal 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

${\displaystyle \ M\ }$

${\displaystyle \ b_{1},\dots ,b_{n}~.}$

${\displaystyle \ b_{1},\dots ,b_{n}\ }$

Uniqueness up to unitary transformations[edit]

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 ${\displaystyle \ M=B^{*}B=B^{*}Q^{*}QB=A^{*}\ }$ 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 ${\displaystyle Q}$isunitary.