Definiteness of a matrix

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

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

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

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

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

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

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

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

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

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

A ${\displaystyle n\times n}$ 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 ${\displaystyle z^{*}Mz}$ is real and positive for all non-zero complex column vectors ${\displaystyle z}$". This condition implies that ${\displaystyle M}$ is Hermitian (i.e. its transpose is equal to its conjugate). To see this, consider the matrices ${\displaystyle A={\tfrac {1}{2}}\left(M+M^{*}\right)}$ and ${\displaystyle B={\tfrac {1}{2i}}\left(M-M^{*}\right)}$, so that ${\displaystyle M=A+iB}$ and ${\displaystyle z^{*}Mz=z^{*}Az+iz^{*}Bz}$. The matrices ${\displaystyle A}$ and ${\displaystyle B}$ are Hermitian, therefore ${\displaystyle z^{*}Az}$ and ${\displaystyle z^{*}Bz}$ are individually real. If ${\displaystyle z^{*}Mz}$ is real, then ${\displaystyle z^{*}Bz}$ must be zero for all ${\displaystyle z}$. Then ${\displaystyle B}$ is the zero matrix and ${\displaystyle M=A}$, proving that ${\displaystyle M}$ is Hermitian.

By this definition, a positive definite real matrix ${\displaystyle M}$ is Hermitian, hence symmetric; and ${\displaystyle z^{\textsf {T}}Mz}$ is positive for all non-zero real column vectors ${\displaystyle z}$. 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 ${\displaystyle z}$ with entries ${\displaystyle a}$ and ${\displaystyle b}$ we have ${\displaystyle z^{\textsf {T}}Mz=(a+b)a+(-a+b)b=a^{2}+b^{2}}$, which is always positive if ${\displaystyle z}$ is not zero. However, if ${\displaystyle z}$ is the complex vector with entries ${\displaystyle 1}$ and ${\displaystyle i}$, one gets

${\displaystyle z^{*}Mz=[1,-i]M[1,i]^{\textsf {T}}=[1+i,1-i][1,i]^{\textsf {T}}=2+2i}$

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

On the other hand, for a symmetric real matrix ${\displaystyle M}$, the condition "${\displaystyle z^{\textsf {T}}Mz>0}$ for all nonzero real vectors ${\displaystyle z}$" 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.

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.

For arbitrary square matrices ${\displaystyle M}$, ${\displaystyle N}$ we write ${\displaystyle M\geq N}$ if ${\displaystyle M-N\geq 0}$ i.e., ${\displaystyle M-N}$ is positive semi-definite. This defines a partial ordering on the set of all square matrices. One can similarly define a strict partial ordering ${\displaystyle M>N}$. The ordering is called the Loewner order.

Every positive definite matrix is invertible and its inverse is also positive definite.[4] If ${\displaystyle M\geq N>0}$ then ${\displaystyle N^{-1}\geq M^{-1}>0}$.[5] Moreover, by the min-max theorem, the kth largest eigenvalue of ${\displaystyle M}$ is greater than the kth largest eigenvalue of ${\displaystyle N}$.

If ${\displaystyle M}$ is positive definite and ${\displaystyle r>0}$ is a real number, then ${\displaystyle rM}$ is positive definite.[6]

If ${\displaystyle M}$ and ${\displaystyle N}$ are positive definite, then the sum ${\displaystyle M+N}$ is also positive definite.[6]

• If ${\displaystyle M}$ and ${\displaystyle N}$ are positive definite, then the products ${\displaystyle MNM}$ and ${\displaystyle NMN}$ are also positive definite. If ${\displaystyle MN=NM}$, then ${\displaystyle MN}$ is also positive definite.

• If ${\displaystyle M}$ is positive semidefinite, then ${\displaystyle Q^{\textsf {T}}MQ}$ is positive semidefinite. If ${\displaystyle M}$ is positive definite and ${\displaystyle Q}$ has full column rank, then ${\displaystyle Q^{\textsf {T}}MQ}$ is positive definite.[7]

For any matrix ${\displaystyle A}$, the matrix ${\displaystyle A^{*}A}$ is positive semidefinite, and ${\displaystyle \operatorname {rank} (A)=\operatorname {rank} (A^{*}A)}$. Conversely, any Hermitian positive semi-definite matrix ${\displaystyle M}$ can be written as ${\displaystyle M=LL^{*}}$, where ${\displaystyle L}$ is lower triangular; this is the Cholesky decomposition. If ${\displaystyle M}$ is not positive definite, then some of the diagonal elements of ${\displaystyle L}$ may be zero.

A hermitian matrix ${\displaystyle M}$ is positive definite if and only if it has a unique Cholesky decomposition, i.e. the matrix ${\displaystyle M}$ is positive definite if and only if there exists a unique lower triangular matrix ${\displaystyle L}$, with real and strictly positive diagonal elements, such that ${\displaystyle M=LL^{*}}$.

A matrix ${\displaystyle M}$ is positive semi-definite if and only if there is a positive semi-definite matrix ${\displaystyle B}$ with ${\displaystyle B^{2}=M}$. This matrix ${\displaystyle B}$ is unique,[8] is called the square root of ${\displaystyle M}$, and is denoted with ${\displaystyle B=M^{\frac {1}{2}}}$ (the square root ${\displaystyle B}$ is not to be confused with the matrix ${\displaystyle L}$ in the Cholesky factorization ${\displaystyle M=LL^{*}}$, which is also sometimes called the square root of ${\displaystyle M}$).

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

Every principal submatrix of a positive definite matrix is positive definite.

The diagonal entries ${\displaystyle m_{ii}}$ of a positive definite matrix are real and non-negative. As a consequence the trace, ${\displaystyle \operatorname {tr} (M)\geq 0}$. Furthermore,[9] since every principal sub-matrix (in particular, 2-by-2) is positive definite,

${\displaystyle \left|m_{ij}\right|\leq {\sqrt {m_{ii}m_{jj}}}\leq {\frac {m_{ii}+m_{jj}}{2}}\quad \forall i,j}$

and thus

${\displaystyle \max _{i,j}\left|m_{ij}\right|\leq \max _{i}\left|m_{ii}\right|}$

An ${\displaystyle n\times n}$ Hermitian matrix ${\displaystyle M}$ is positive definite if it satisfies the following trace inequalities:[10]

${\displaystyle \operatorname {tr} (M)>0\quad \mathrm {and} \quad {\frac {(\operatorname {tr} (M))^{2}}{\operatorname {tr} (M^{2})}}>n-1.}$

If ${\displaystyle M,N\geq 0}$, although ${\displaystyle MN}$ is not necessary positive semidefinite, the Hadamard product ${\displaystyle M\circ N\geq 0}$ (this result is often called the Schur product theorem).[11]

Regarding the Hadamard product of two positive semidefinite matrices ${\displaystyle M=(m_{ij})\geq 0}$, ${\displaystyle N\geq 0}$, there are two notable inequalities:

• Oppenheim's inequality: ${\displaystyle \det(M\circ N)\geq \det(N)\prod \nolimits _{i}m_{ii}.}$[12]
• ${\displaystyle \det(M\circ N)\geq \det(M)\det(N)}$.[13]

If ${\displaystyle M,N\geq 0}$, although ${\displaystyle MN}$ is not necessary positive semidefinite, the Kronecker product ${\displaystyle M\otimes N\geq 0}$.

If ${\displaystyle M,N\geq 0}$, although ${\displaystyle MN}$ is not necessary positive semidefinite, the Frobenius product ${\displaystyle M:N\geq 0}$ (Lancaster–Tismenetsky, The Theory of Matrices, p. 218).

The set of positive semidefinite symmetric matrices is convex. That is, if ${\displaystyle M}$ and ${\displaystyle N}$ are positive semidefinite, then for any ${\displaystyle \alpha }$ between 0 and 1, ${\displaystyle \alpha M+(1-\alpha )N}$ is also positive semidefinite. For any vector ${\displaystyle x}$:

${\displaystyle x^{\textsf {T}}\left(\alpha M+(1-\alpha )N\right)x=\alpha x^{\textsf {T}}Mx+(1-\alpha )x^{\textsf {T}}Nx\geq 0.}$

This property guarantees that semidefinite programming problems converge to a globally optimal solution.

The positive-definiteness of a matrix ${\displaystyle A}$ expresses that the angle ${\displaystyle \theta }$ between any vector ${\displaystyle x}$ and its image ${\displaystyle Ax}$ is always ${\displaystyle -\pi /2<\theta <+\pi /2}$:

${\displaystyle \cos(\theta )={\frac {x^{T}Ax}{\left\Vert x\right\Vert \left\Vert Ax\right\Vert }}={\frac {\langle x,Ax\rangle }{\left\Vert x\right\Vert \left\Vert Ax\right\Vert }},\theta =\theta (x,Ax)={\widehat {x,Ax}}=the\ angle\ between\ x\ and\ Ax}$

A Hermitian matrix is positive semidefinite if and only if all of its principal minors are nonnegative. It is however not enough to consider the leading principal minors only, as is checked on the diagonal matrix with entries 0 and −1.

A positive ${\displaystyle 2n\times 2n}$ matrix may also be defined by blocks:

${\displaystyle M={\begin{bmatrix}A&B\\C&D\end{bmatrix}}}$

where each block is ${\displaystyle n\times n}$. By applying the positivity condition, it immediately follows that ${\displaystyle A}$ and ${\displaystyle D}$ are hermitian, and ${\displaystyle C=B^{*}}$.

We have that ${\displaystyle z^{*}Mz\geq 0}$ for all complex ${\displaystyle z}$, and in particular for ${\displaystyle z=[v,0]^{\textsf {T}}}$. Then

${\displaystyle {\begin{bmatrix}v^{*}&0\end{bmatrix}}{\begin{bmatrix}A&B\\B^{*}&D\end{bmatrix}}{\begin{bmatrix}v\\0\end{bmatrix}}=v^{*}Av\geq 0.}$

A similar argument can be applied to ${\displaystyle D}$, and thus we conclude that both ${\displaystyle A}$ and ${\displaystyle D}$ must be positive definite matrices, as well.

Converse results can be proved with stronger conditions on the blocks, for instance using the Schur complement.

Fourier's law of heat conduction, giving heat flux ${\displaystyle q}$ in terms of the temperature gradient ${\displaystyle g=\nabla T}$ is written for anisotropic media as ${\displaystyle q=-Kg}$, in which ${\displaystyle K}$ is the symmetric thermal conductivity matrix. The negative is inserted in Fourier's law to reflect the expectation that heat will always flow from hot to cold. In other words, since the temperature gradient ${\displaystyle g}$ always points from cold to hot, the heat flux ${\displaystyle q}$ is expected to have a negative inner product with ${\displaystyle g}$ so that ${\displaystyle q^{\textsf {T}}g<0}$. Substituting Fourier's law then gives this expectation as ${\displaystyle g^{\textsf {T}}Kg>0}$, implying that the conductivity matrix should be positive definite.