Inmathematics, a symmetric matrix${\displaystyle M}$ withreal entries ispositive-definite if the real number${\displaystyle {\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}}$ is positive for every nonzero realcolumn vector${\displaystyle \mathbf{x},}$ where${\displaystyle {\mathbf{x}}^{\mathrm{\top }}}$ is therow vectortranspose of${\displaystyle \mathbf{x}.}$^[1]More generally, aHermitian matrix (that is, acomplex matrix equal to itsconjugate transpose) ispositive-definite if the real number${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}}$ is positive for every nonzero complex column vector${\displaystyle \mathbf{z},}$ where${\displaystyle {\mathbf{z}}^{\ast }}$ denotes the conjugate transpose of${\displaystyle \mathbf{z}.}$

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

Some authors use more general definitions of definiteness, permitting the matrices to be non-symmetric or non-Hermitian. The properties of these generalized definite matrices are explored in§ Extension for non-Hermitian square matrices, below, but are not the main focus of this article.

It follows from the above definitions that a matrix is positive-definiteif and only if it is the matrix of apositive-definite quadratic form orHermitian form. In other words, a matrix is positive-definite if and only if it defines aninner 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 matrixM is positive-definite if and only if it satisfies any of the following equivalent conditions.

• ${\displaystyle M}$ iscongruent with adiagonal matrix with positive real entries.
• ${\displaystyle M}$ is symmetric or Hermitian, and all itseigenvalues are real and positive.
• ${\displaystyle M}$ is symmetric or Hermitian, and all its leadingprincipal minors are positive.
• There exists aninvertible matrix${\displaystyle B}$ with conjugate transpose${\displaystyle B^{\ast }}$ such that${\displaystyle M=B^{\ast }B.}$

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 ofconvex optimization, since, given afunction of several real variables that is twicedifferentiable, then if itsHessian matrix (matrix of its second partial derivatives) is positive-definite at a point${\displaystyle p,}$ then the function isconvex nearp, 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 anopenconvex cone, while the set of positive semi-definite matrices is aclosed convex cone.^[2]

In the following definitions,${\displaystyle {\mathbf{x}}^{\mathrm{\top }}}$ is the transpose of${\displaystyle \mathbf{x},}$${\displaystyle {\mathbf{z}}^{\ast }}$ is theconjugate transpose of${\displaystyle \mathbf{z},}$ and${\displaystyle \mathbf{0}}$ denotes then dimensional zero-vector.

An${\displaystyle n\times n}$ symmetric real matrix${\displaystyle M}$ is said to bepositive-definite if${\displaystyle {\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}>0}$ 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 bepositive-semidefinite ornon-negative-definite if${\displaystyle {\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}\ge 0}$ for all${\displaystyle \mathbf{x}}$ in${\displaystyle {\mathbb{R}}^n.}$ Formally,

An${\displaystyle n\times n}$ symmetric real matrix${\displaystyle M}$ is said to benegative-definite if 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 benegative-semidefinite ornon-positive-definite if${\displaystyle {\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}\le 0}$ 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 calledindefinite.

The following definitions all involve the term${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}.}$ 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 bepositive-definite if${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}>0}$ 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 bepositive semi-definite ornon-negative-definite if${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}\ge 0}$ for all${\displaystyle \mathbf{z}}$ in${\displaystyle {\mathbb{C}}^n.}$ Formally,

An${\displaystyle n\times n}$ Hermitian complex matrix${\displaystyle M}$ is said to benegative-definite if 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 benegative semi-definite ornon-positive-definite if${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}\le 0}$ 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 calledindefinite.

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${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}}$ 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${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}}$ being real, it equals its conjugate transpose${\displaystyle {\mathbf{z}}^{\ast }{M}^{\ast }\mathbf{z}}$ for every${\displaystyle \mathbf{z},}$ which implies${\displaystyle M=M^{\ast }.}$

By this definition, a positive-definitereal matrix${\displaystyle M}$ is Hermitian, hence symmetric; and${\displaystyle {\mathbf{z}}^{\mathrm{\top }}M\mathbf{z}}$ is positive for all non-zeroreal 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${\displaystyle {\mathbf{z}}^{\mathrm{\top }}M\mathbf{z}=\left(a+b\right)a+\left(-a+b\right)b=a^2+b^2,}$ which is always positive if${\displaystyle \mathbf{z}}$ is not zero. However, if${\displaystyle \mathbf{z}}$ is the complex vector with entries1 and${\displaystyle i,}$ one gets

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

On the other hand, for asymmetric real matrix${\displaystyle M,}$ the condition "${\displaystyle {\mathbf{z}}^{\mathrm{\top }}M\mathbf{z}>0}$ 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${\displaystyle M⪰0}$ and if${\displaystyle M}$ is positive-definite one writes${\displaystyle M\succ 0.}$ To denote that${\displaystyle M}$ is negative semi-definite one writes${\displaystyle M⪯0}$ and to denote that${\displaystyle M}$ is negative-definite one writes${\displaystyle M\prec 0.}$

The notion comes fromfunctional analysis where positive semidefinite matrices definepositive operators. If two matrices${\displaystyle A}$ and${\displaystyle B}$ satisfy${\displaystyle B-A⪰0,}$ we can define anon-strict partial order${\displaystyle B⪰A}$ that isreflexive,antisymmetric, andtransitive; It is not atotal order, however, as${\displaystyle B-A,}$ in general, may be indefinite.

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

Let${\displaystyle M}$ be an${\displaystyle n\times n}$Hermitian matrix (this includes realsymmetric 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${\displaystyle PD{P}^{-1}}$ be aneigendecomposition of${\displaystyle M,}$ where${\displaystyle P}$ is aunitary complex matrix whose columns comprise anorthonormal basis ofeigenvectors of${\displaystyle M,}$ and${\displaystyle D}$ is arealdiagonal matrix whosemain diagonal contains the correspondingeigenvalues. 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 aschanging the basis to the eigenvector coordinate system using${\displaystyle P^{-1},}$ giving${\displaystyle P^{-1}\mathbf{z},}$ applying thestretching transformation${\displaystyle D}$ to the result, giving${\displaystyle D{P}^{-1}\mathbf{z},}$ and then changing the basis back using${\displaystyle P,}$ giving${\displaystyle PD{P}^{-1}\mathbf{z}.}$

With this in mind, the one-to-one change of variable${\displaystyle \mathbf{y}=P\mathbf{z}}$ shows that${\displaystyle {\mathbf{z}}^{\ast }M\mathbf{z}}$ is real and positive for any complex vector${\displaystyle \mathbf{z}}$ if and only if${\displaystyle {\mathbf{y}}^{\ast }D\mathbf{y}}$ 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 thespectral theorem guarantees all eigenvalues of a Hermitian matrix to be real, the positivity of eigenvalues can be checked usingDescartes' rule of alternating signs when thecharacteristic 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$${\displaystyle M=B^{\ast }B}$$of a matrix${\displaystyle B}$ with itsconjugate transpose.

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

${\displaystyle M}$ 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^{\ast }B,}$${\displaystyle \mathrm{rank}(M)=\mathrm{rank}(B).}$^[3]

The columns${\displaystyle b_1,\dots ,b_n}$ of${\displaystyle B}$ can be seen as vectors in thecomplex orreal vector space${\displaystyle {\mathbb{R}}^k,}$ respectively.Then the entries of${\displaystyle M}$ areinner products (that isdot products, in the real case) of these vectors$${\displaystyle M_{ij}=⟨b_i,b_j⟩.}$$In other words, a Hermitian matrix${\displaystyle M}$ is positive semidefinite if and only if it is theGram matrix of some vectors${\displaystyle b_1,\dots ,b_n.}$It is positive definite if and only if it is the Gram matrix of somelinearly independent vectors.In general, the rank of the Gram matrix of vectors${\displaystyle b_1,\dots ,b_n}$ equals the dimension of the spacespanned by these vectors.^[4]

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

However, this is the only way in which two decompositions can differ: The decomposition is unique up tounitary 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${\displaystyle A^{\ast }A=B^{\ast }B,}$then there is a${\displaystyle \ell \times k}$ matrix${\displaystyle Q}$ with orthonormal columns (meaning${\displaystyle Q^{\ast }Q=I_{k\times k}}$) such that${\displaystyle B=QA.}$^[5]When${\displaystyle \ell =k}$ this means${\displaystyle Q}$ isunitary.

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 anorthogonal matrix, which describes arigid transformation (an isometry of Euclidean space${\displaystyle {\mathbb{R}}^k}$) preserving the 0 point (i.e.rotations andreflections, 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${\displaystyle B^{\ast }=B}$) satisfying${\displaystyle M=BB.}$ This matrix${\displaystyle B}$ is unique,^[6] is called thenon-negativesquare 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 thepositive square root of${\displaystyle M.}$

The non-negative square root should not be confused with other decompositions${\displaystyle M=B^{\ast }B.}$Some authors use the namesquare root and${\displaystyle M^{\frac{1}{2}}}$ for any such decomposition, or specifically for theCholesky 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${\displaystyle M=L{L}^{\ast },}$ where${\displaystyle L}$ is lower triangular with non-negative diagonal (equivalently${\displaystyle M=B^{\ast }B}$ where${\displaystyle B=L^{\ast }}$ is upper triangular); this is theCholesky 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${\displaystyle L{L}^{\ast }}$ is positive semidefinite. The Cholesky decomposition is especially useful for efficient numerical calculations.A closely related decomposition is theLDL decomposition,${\displaystyle M=LD{L}^{\ast },}$ where${\displaystyle D}$ is diagonal and${\displaystyle L}$ islower 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${\displaystyle S\in \mathbf{S}\mathbf{p}(2n,\mathbb{R})}$ and diagonal real positive${\displaystyle D\in {\mathbb{R}}^{n\times n}}$ such that${\displaystyle SM{S}^T=D\oplus D}$.

Let${\displaystyle M}$ be an${\displaystyle n\times n}$real symmetric matrix, and let${\displaystyle B_1(M)\equiv \{\mathbf{x}\in {\mathbb{R}}^n:{\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}\le 1\}}$ be the "unit ball" defined by${\displaystyle M.}$ Then we have the following

• ${\displaystyle B_1(\mathbf{v}{\mathbf{v}}^{\mathrm{\top }})}$ is a solid slab sandwiched between${\displaystyle \pm \{\mathbf{w}:⟨\mathbf{w},\mathbf{v}⟩=1\}.}$
• ${\displaystyle M⪰0}$ if and only if${\displaystyle B_1(M)}$ is an ellipsoid, or an ellipsoidal cylinder.
• ${\displaystyle M\succ 0}$ if and only if${\displaystyle B_1(M)}$ is bounded, that is, it is an ellipsoid.
• If${\displaystyle N\succ 0,}$ then${\displaystyle M⪰N}$ if and only if${\displaystyle B_1(M)\subseteq B_1(N);}$${\displaystyle M\succ N}$ if and only if${\displaystyle B_1(M)\subseteq \mathrm{int}(B_1(N)).}$
• If${\displaystyle N\succ 0,}$ then${\displaystyle M⪰\frac{\mathbf{v}{\mathbf{v}}^{\mathrm{\top }}}{{\mathbf{v}}^{\mathrm{\top }}N\mathbf{v}}}$ for all${\displaystyle v\ne 0}$ if and only if$B_1(M)\subset \bigcap _{{\mathbf{v}}^{\mathrm{\top }}N\mathbf{v}=1}{B}_1(\mathbf{v}{\mathbf{v}}^{\mathrm{\top }}).$ So, since the polar dual of an ellipsoid is also an ellipsoid with the same principal axes, with inverse lengths, we have$${\displaystyle B_1(N^{-1})=\bigcap _{{\mathbf{v}}^{\mathrm{\top }}N\mathbf{v}=1}{B}_1(\mathbf{v}{\mathbf{v}}^{\mathrm{\top }})=\bigcap _{{\mathbf{v}}^{\mathrm{\top }}N\mathbf{v}=1}\{\mathbf{w}:|⟨\mathbf{w},\mathbf{v}⟩|\le 1\}.}$$ That is, if${\displaystyle N}$ is positive-definite, then${\displaystyle M⪰\frac{\mathbf{v}{\mathbf{v}}^{\mathrm{\top }}}{{\mathbf{v}}^{\mathrm{\top }}N\mathbf{v}}}$ for all${\displaystyle \mathbf{v}\ne \mathbf{0}}$ if and only if${\displaystyle M⪰N^{-1}.}$

Let${\displaystyle M}$ be an${\displaystyle n\times n}$Hermitian matrix. The following properties are equivalent to${\displaystyle M}$ being positive definite:

The associated sesquilinear form is an inner product

Thesesquilinear form defined by${\displaystyle M}$ is the function${\displaystyle ⟨\cdot ,\cdot ⟩}$ from${\displaystyle {\mathbb{C}}^n\times {\mathbb{C}}^n}$ to${\displaystyle {\mathbb{C}}^n}$ such that${\displaystyle ⟨\mathbf{x},\mathbf{y}⟩\equiv {\mathbf{y}}^{\ast }M\mathbf{x}}$ for all${\displaystyle \mathbf{x}}$ and${\displaystyle \mathbf{y}}$ in${\displaystyle {\mathbb{C}}^n,}$ where${\displaystyle {\mathbf{y}}^{\ast }}$ is the conjugate transpose of${\displaystyle \mathbf{y}.}$ For any complex matrix${\displaystyle M,}$ this form is linear in${\displaystyle x}$ and semilinear in${\displaystyle \mathbf{y}.}$ Therefore, the form is aninner product on${\displaystyle {\mathbb{C}}^n}$ if and only if${\displaystyle ⟨\mathbf{z},\mathbf{z}⟩}$ is real and positive for all nonzero${\displaystyle \mathbf{z};}$ that is if and only if${\displaystyle M}$ is positive definite. (In fact, every inner product on${\displaystyle {\mathbb{C}}^n}$ arises in this fashion from a Hermitian positive definite matrix.)

Its leading principal minors are all positive

Thekthleading principal minor of a matrix${\displaystyle M}$ is thedeterminant of its upper-left${\displaystyle k\times k}$ sub-matrix. It turns out that a matrix is positive definite if and only if all these determinants are positive. This condition is known asSylvester's criterion, and provides an efficient test of positive definiteness of a symmetric real matrix. Namely, the matrix is reduced to anupper triangular matrix by usingelementary row operations, as in the first part of theGaussian elimination method, taking care to preserve the sign of its determinant duringpivoting process. Since thekth leading principal minor of a triangular matrix is the product of its diagonal elements up to row${\displaystyle k,}$ Sylvester's criterion is equivalent to checking whether its diagonal elements are all positive. This condition can be checked each time a new row${\displaystyle k}$ of the triangular matrix is obtained.

A positive semidefinite matrix is positive definite if and only if it isinvertible.^[7]A matrix${\displaystyle M}$ is negative (semi)definite if and only if${\displaystyle -M}$ is positive (semi)definite.

The (purely)quadratic form associated with a real${\displaystyle n\times n}$ matrix${\displaystyle M}$ is the function${\displaystyle Q:{\mathbb{R}}^n\to \mathbb{R}}$ such that${\displaystyle Q(\mathbf{x})={\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}}$ for all${\displaystyle \mathbf{x}.}$${\displaystyle M}$ can be assumed symmetric by replacing it with${\displaystyle \frac{1}{2}\left(M+M^{\mathrm{\top }}\right),}$ since any asymmetric part will be zeroed-out in the double-sided product.

A symmetric matrix${\displaystyle M}$ is positive definite if and only if its quadratic form is astrictly convex function.

More generally, anyquadratic function from${\displaystyle {\mathbb{R}}^n}$ to${\displaystyle \mathbb{R}}$ can be written as${\displaystyle {\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}+{\mathbf{b}}^{\mathrm{\top }}\mathbf{x}+c}$ where${\displaystyle M}$ is a symmetric${\displaystyle n\times n}$ matrix,${\displaystyle \mathbf{b}}$ is a realn vector, and${\displaystyle c}$ a real constant. In the${\displaystyle n=1}$ case, this is a parabola, and just like in the${\displaystyle n=1}$ case, we have

Theorem: This quadratic function is strictly convex, and hence has a unique finite global minimum, if and only if${\displaystyle M}$ is positive definite.

Proof: If${\displaystyle M}$ is positive definite, then the function is strictly convex. Its gradient is zero at the unique point of${\displaystyle M^{-1}\mathbf{b},}$ which must be the global minimum since the function is strictly convex. If${\displaystyle M}$ is not positive definite, then there exists some vector${\displaystyle \mathbf{v}}$ such that${\displaystyle {\mathbf{v}}^{\mathrm{\top }}M\mathbf{v}\le 0,}$ so the function${\displaystyle f(t)\equiv (t\mathbf{v}{)}^{\mathrm{\top }}M(t\mathbf{v})+b^{\mathrm{\top }}(t\mathbf{v})+c}$ is a line or a downward parabola, thus not strictly convex and not having a global minimum.

For this reason, positive definite matrices play an important role inoptimization problems.

One symmetric matrix and another matrix that is both symmetric and positive definite can besimultaneously diagonalized. This is so although simultaneous diagonalization is not necessarily performed with asimilarity transformation. This result does not extend to the case of three or more matrices. In this section we write for the real case. Extension to the complex case is immediate.

Let${\displaystyle M}$ be a symmetric and${\displaystyle N}$ a symmetric and positive definite matrix. Write the generalized eigenvalue equation as${\displaystyle \left(M-\lambda N\right)\mathbf{x}=0}$ where we impose that${\displaystyle \mathbf{x}}$ be normalized, i.e.${\displaystyle {\mathbf{x}}^{\mathrm{\top }}N\mathbf{x}=1.}$ Now we useCholesky decomposition to write the inverse of${\displaystyle N}$ as${\displaystyle Q^{\mathrm{\top }}Q.}$ Multiplying by${\displaystyle Q}$ and letting${\displaystyle \mathbf{x}=Q^{\mathrm{\top }}\mathbf{y},}$ we get${\displaystyle Q\left(M-\lambda N\right)Q^{\mathrm{\top }}\mathbf{y}=0,}$ which can be rewritten as${\displaystyle \left(QM{Q}^{\mathrm{\top }}\right)\mathbf{y}=\lambda \mathbf{y}}$ where${\displaystyle {\mathbf{y}}^{\mathrm{\top }}\mathbf{y}=1.}$ Manipulation now yields${\displaystyle MX=NX\mathrm{\Lambda }}$ where${\displaystyle X}$ is a matrix having as columns the generalized eigenvectors and${\displaystyle \mathrm{\Lambda }}$ is a diagonal matrix of the generalized eigenvalues. Now premultiplication with${\displaystyle X^{\mathrm{\top }}}$ gives the final result:${\displaystyle X^{\mathrm{\top }}MX=\mathrm{\Lambda }}$ and${\displaystyle X^{\mathrm{\top }}NX=I,}$ but note that this is no longer an orthogonal diagonalization with respect to the inner product where${\displaystyle {\mathbf{y}}^{\mathrm{\top }}\mathbf{y}=1.}$ In fact, we diagonalized${\displaystyle M}$ with respect to the inner product induced by${\displaystyle N.}$^[8]

Note that this result does not contradict what is said on simultaneous diagonalization in the articleDiagonalizable matrix, which refers to simultaneous diagonalization by a similarity transformation. Our result here is more akin to a simultaneous diagonalization of two quadratic forms, and is useful for optimization of one form under conditions on the other.

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

Every positive definite matrix isinvertible and its inverse is also positive definite.^[9] If${\displaystyle M\ge N>0}$ then${\displaystyle N^{-1}\ge M^{-1}>0.}$^[10] Moreover, by themin-max theorem, thekth largest eigenvalue of${\displaystyle M}$ is greater than or equal to thekth 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.^[11]

• If${\displaystyle M}$ and${\displaystyle N}$ are positive-definite, then the sum${\displaystyle M+N}$ is also positive-definite.^[11]
• If${\displaystyle M}$ and${\displaystyle N}$ are positive-semidefinite, then the sum${\displaystyle M+N}$ is also positive-semidefinite.
• If${\displaystyle M}$ is positive-definite and${\displaystyle N}$ is positive-semidefinite, then the sum${\displaystyle M+N}$ is also positive-definite.

• 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${\displaystyle A^{\ast }MA}$ is positive semidefinite for any (possibly rectangular) matrix${\displaystyle A.}$ If${\displaystyle M}$ is positive definite and${\displaystyle A}$ has full column rank, then${\displaystyle A^{\ast }MA}$ is positive definite.^[12]

The diagonal entries${\displaystyle m_{ii}}$ of a positive-semidefinite matrix are real and non-negative. As a consequence thetrace,${\displaystyle \mathrm{tr}(M)\ge 0.}$ Furthermore,^[13] since every principal sub-matrix (in particular, 2-by-2) is positive semidefinite,$${\displaystyle \left|m_{ij}\right|\le \sqrt{m_{ii}{m}_{jj}}\mathrm{\forall }i,j}$$

and thus, when${\displaystyle n\ge 1,}$$${\displaystyle \underset{i,j}{max}\left|m_{ij}\right|\le \underset{i}{max}{m}_{ii}}$$

An${\displaystyle n\times n}$ Hermitian matrix${\displaystyle M}$ is positive definite if it satisfies the following trace inequalities:^[14]$${\displaystyle \mathrm{tr}(M)>0\mathrm{a}\mathrm{n}\mathrm{d}\frac{(\mathrm{tr}(M){)}^2}{\mathrm{tr}(M^2)}>n-1.}$$

Another important result is that for any${\displaystyle M}$ and${\displaystyle N}$ positive-semidefinite matrices,${\displaystyle \mathrm{tr}(MN)\ge 0.}$ This follows by writing${\displaystyle \mathrm{tr}(MN)=\mathrm{tr}(M^{\frac{1}{2}}N{M}^{\frac{1}{2}}).}$ The matrix${\displaystyle M^{\frac{1}{2}}N{M}^{\frac{1}{2}}}$ is positive-semidefinite and thus has non-negative eigenvalues, whose sum, the trace, is therefore also non-negative.

If${\displaystyle M,N\ge 0,}$ although${\displaystyle MN}$ is not necessary positive semidefinite, theHadamard product is,${\displaystyle M\circ N\ge 0}$ (this result is often called theSchur product theorem).^[15]

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

• Oppenheim's inequality:${\displaystyle det(M\circ N)\ge det(N){\prod }_i{m}_{ii}.}$^[16]
• ${\displaystyle det(M\circ N)\ge det(M)det(N).}$^[17]

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

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

The set of positive semidefinite symmetric matrices isconvex. That is, if${\displaystyle M}$ and${\displaystyle N}$ are positive semidefinite, then for any${\displaystyle \alpha }$ between0 and1,${\displaystyle \alpha M+\left(1-\alpha \right)N}$ is also positive semidefinite. For any vector${\displaystyle \mathbf{x}}$:$${\displaystyle {\mathbf{x}}^{\mathrm{\top }}\left(\alpha M+\left(1-\alpha \right)N\right)\mathbf{x}=\alpha {\mathbf{x}}^{\mathrm{\top }}M\mathbf{x}+(1-\alpha ){\mathbf{x}}^{\mathrm{\top }}N\mathbf{x}\ge 0.}$$

This property guarantees thatsemidefinite 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${\displaystyle \mathbf{x}}$ and its image${\displaystyle A\mathbf{x}}$ is always

$${\displaystyle \cos \theta =\frac{{\mathbf{x}}^{\mathrm{\top }}A\mathbf{x}}{\Vert \mathbf{x}\Vert \Vert A\mathbf{x}\Vert }=\frac{⟨\mathbf{x},A\mathbf{x}⟩}{\Vert \mathbf{x}\Vert \Vert A\mathbf{x}\Vert },\theta =\theta (\mathbf{x},A\mathbf{x})\equiv \widehat{\left(\mathbf{x},A\mathbf{x}\right)}\equiv }$$ the angle between${\displaystyle \mathbf{x}}$ and${\displaystyle A\mathbf{x}.}$

1. If${\displaystyle M}$ is a symmetricToeplitz matrix, i.e. the entries${\displaystyle m_{ij}}$ are given as a function of their absolute index differences:${\displaystyle m_{ij}=h(|i-j|),}$ and thestrict inequality holds, then${\displaystyle M}$ isstrictly positive definite.
2. Let${\displaystyle M>0}$ and${\displaystyle N}$ Hermitian. If${\displaystyle MN+NM\ge 0}$ (resp.,${\displaystyle MN+NM>0}$) then${\displaystyle N\ge 0}$ (resp.,${\displaystyle N>0}$).^[18]
3. If${\displaystyle M>0}$ is real, then there is a${\displaystyle \delta >0}$ such that${\displaystyle M>\delta I,}$ where${\displaystyle I}$ is theidentity matrix.
4. If${\displaystyle M_k}$ denotes the leading${\displaystyle k\times k}$ minor,${\displaystyle det\left(M_k\right)/det\left(M_{k-1}\right)}$ is thekth pivot duringLU decomposition.
5. A matrix is negative definite if itskth order leadingprincipal minor is negative when${\displaystyle k}$ is odd, and positive when${\displaystyle k}$ is even.
6. If${\displaystyle M}$ is a real positive definite matrix, then there exists a positive real number${\displaystyle m}$ such that for every vector${\displaystyle \mathbf{v},}$${\displaystyle {\mathbf{v}}^{\mathrm{\top }}M\mathbf{v}\ge m\Vert \mathbf{v}{\Vert }_2^2.}$
7. 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 entries0 and−1 .

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