Definite matrix

In linear algebra, a positive-definite matrix is a matrix which in many ways is analogous to a positive real number. The notion is closely related to a positive-definite symmetric bilinear form (or a sesquilinear form in the complex case).

The proper definition of positive-definite is unambiguous for Hermitian matrices, but there is no agreement in the literature on how this should be extended for non-Hermitian matrices, if at all. (See the section Non-Hermitian matrices below.)

Definition

An n × n real symmetric matrix M is positive definite if z^TMz > 0 for all non-zero vectors z with real entries ($z\in \mathbb {R} ^{n}$), where z^T denotes the transpose of z.

An n × n complex Hermitian matrix M is positive definite if z^*Mz > 0 for all non-zero complex vectors z, where z^* denotes the conjugate transpose of z. The quantity z^*Mz is always real because M is a Hermitian matrix.

Examples

The matrix $M_{0}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ is positive definite. For a vector with entries ${\textbf {z}}={\begin{bmatrix}z_{0}\\z_{1}\end{bmatrix}}$ the quadratic form is ${\begin{bmatrix}z_{0}&z_{1}\end{bmatrix}}{\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{bmatrix}z_{0}\\z_{1}\end{bmatrix}}=z_{0}^{2}+z_{1}^{2};$ when the entries z₀, z₁ are real and at least one of them nonzero, this is positive.

The matrix $M_{1}={\begin{bmatrix}0&1\\1&0\end{bmatrix}}$ is not positive definite. When ${\textbf {z}}={\begin{bmatrix}1\\-1\end{bmatrix}}$ the quadratic form at z is then

${\begin{bmatrix}1&-1\end{bmatrix}}{\begin{bmatrix}0&1\\1&0\end{bmatrix}}{\begin{bmatrix}1\\-1\end{bmatrix}}=-2<0.$

Another example of positive definite matrix is given by

$A={\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}}.$

It is positive definite since for any non-zero vector $x={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$, we have

$x^{t}Ax={\begin{bmatrix}x_{1}&x_{2}&x_{3}\end{bmatrix}}{\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$

$={\begin{bmatrix}(2x_{1}-x_{2})&(-x_{1}+2x_{2}-x_{3})&(-x_{2}+2x_{3})\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$

$=2x_{1}^{2}-2x_{1}x_{2}+2x_{2}^{2}-2x_{2}x_{3}+2x_{3}^{2}$

$=x_{1}^{2}+(x_{1}-x_{2})^{2}+(x_{2}-x_{3})^{2}+x_{3}^{2}>0.$

For a huge class of examples, consider that in statistics positive definite matrices appear as covariance matrices. In fact all positive definite matrices are a covariance matrix for some probability distribution.

Characterizations

Let M be an n × n Hermitian matrix. The following properties are equivalent to M being positive definite:

1.	All eigenvalues $\lambda _{i}$ of $M$ are positive. Recall that any Hermitian M, by the spectral theorem, may be regarded as a real diagonal matrix D that has been re-expressed in some new coordinate system (i.e., $M=P^{-1}DP$ for some unitary matrix P whose rows are orthonormal eigenvectors of M, forming a basis). So this characterization means that M is positive definite if and only if the diagonal elements of D (the eigenvalues) are all positive. In other words, in the basis consisting of the eigenvectors of M, the action of M is component-wise multiplication with a (fixed) element in Cn with positive entries.
2.	The sesquilinear form $\langle \mathbf {x} ,\mathbf {y} \rangle :=\mathbf {y} ^{*}\mathbf {M} \mathbf {x}$ defines an inner product on Cn. (In fact, every inner product on Cn arises in this fashion from a Hermitian positive definite matrix.)
3.	M is the Gram matrix of some collection of linearly independent vectors ${\textbf {x}}_{1},\ldots ,{\textbf {x}}_{n}\in \mathbb {C} ^{k}$ for some k. That is, M satisfies: $M_{ij}=\langle {\textbf {x}}_{i},{\textbf {x}}_{j}\rangle ={\textbf {x}}_{i}^{*}{\textbf {x}}_{j}.$ The vectors xi may optionally be restricted to fall in Cn. In other words, M is of the form $A^{*}A$ where A is not necessarily square but must be injective in general.
4.	All the following matrices have a positive determinant (Sylvester's criterion): the upper left 1-by-1 corner of $M$ the upper left 2-by-2 corner of $M$ the upper left 3-by-3 corner of $M$ ... $M$ itself In other words, all of the leading principal minors are positive. For positive semidefinite matrices, all principal minors have to be non-negative. The leading principal minors alone do not imply positive semidefiniteness, as can be seen from the example ${\begin{bmatrix}1&1&1\\1&1&1\\1&1&-1\end{bmatrix}}.$
5.	There exists a unique lower triangular matrix $L$, with strictly positive diagonal elements, that allows the factorization of $M$ into $M=LL^{*}$. where $L^{*}$ is the conjugate transpose of $L$. This factorization is called Cholesky decomposition.
6.	The quadratic form associated with M $f(\mathbf {x} )={\frac {1}{2}}\mathbf {x} ^{\mathrm {T} }M\mathbf {x} -\mathbf {x} ^{\mathrm {T} }\mathbf {b}$ is, regardless of b, a convex function. $f(\mathbf {x} )$ has a global minimum, and this is why positive definite matrices are so common in optimization problems.

For real symmetric matrices, these properties can be simplified by replacing $\mathbb {C} ^{n}$ with $\mathbb {R} ^{n}$, and "conjugate transpose" with "transpose."

Quadratic forms

Echoing condition 2 above, one can also formulate positive-definiteness in terms of quadratic forms. Let K be the field R or C, and V be a vector space over K. A Hermitian form

$B:V\times V\rightarrow K$

is a bilinear map such that B(x, y) is always the complex conjugate of B(y, x). Such a function B is called positive definite if B(x, x) > 0 for every nonzero x in V.

Simultaneous diagonalization

Two symmetric, positive-definite matrices can be simultaneously diagonalized, although not necessarily via a similarity. 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 $M$ and $N$ be two positive-definite matrices. Write the generalized eigenvalue equation as $(M-\lambda N)x=0$ where $x^{T}Nx=1$. Now we can decompose the inverse of $N$ as $QQ^{T}$ (so $N$ _must_ be positive definite, as the proof shows, in fact it is enough that $M$ is symmetric). Now multiply varios places with $Q$ to get $Q(M-\lambda N)Q^{T}Q^{-T}x=0$ which we can rewrite as $(QMQ^{T})y=\lambda y$ where $y^{T}y=1$. Manipulation now yields $MX=NX\Lambda$ where $X$ is a matrix having as columns the generalized eigenvectors and $\Lambda$ is a diagonal matrix with the generalized eigenvalues. Now premultiplication with $X^{T}$ gives the final result:

$X^{T}MX=\Lambda$ and $X^{T}NX=I$, but note that this is no longer an orthogonal diagonalization.

Note that this result does not contradict what is said on simultaneous diagonalization in the article Diagonalizable matrix#Simultaneous diagonalisation, 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 this result see Horn&Johnson, 1985, page 218 and following.

Negative-definite, semidefinite and indefinite matrices

The n × n Hermitian matrix M is said to be negative-definite if

$x^{*}Mx<0\,$

for all non-zero $x\in \mathbb {C} ^{n}$ (or, all non-zero $x\in \mathbb {R} ^{n}$ for the real matrix).

It is called positive-semidefinite (or sometimes nonnegative-definite) if

$x^{*}Mx\geq 0$

for all $x\in \mathbb {C} ^{n}$ (or, all $x\in \mathbb {R} ^{n}$ for the real matrix), where $x^{*}$ is the conjugate transpose of x.

It is called negative-semidefinite if

$x^{*}Mx\leq 0$

for all $x\in \mathbb {C} ^{n}$ (or, all $x\in \mathbb {R} ^{n}$ for the real matrix).

A matrix M is positive-semidefinite if and only if it arises as the Gram matrix of some set of vectors. In contrast to the positive-definite case, these vectors need not be linearly independent.

For any matrix A, the matrix A^*A is positive semidefinite, and rank(A) = rank(A^*A). Conversely, any Hermitian positive semidefinite matrix M can be written as M = A^*A; this is the Cholesky decomposition.

A Hermitian matrix which is neither positive- nor negative-semidefinite is called indefinite.

A matrix is negative definite if all kth order leading principal minors are negative if k is odd and positive if k is even.

A Hermitian matrix is negative-definite, negative-semidefinite, or positive-semidefinite if and only if all of its eigenvalues are negative, non-positive, or non-negative, respectively.

Further properties

If M is positive-semidefinite, one sometimes writes M ≥ 0 and if M is positive-definite one writes M > 0. The notion comes from functional analysis where positive definite matrices define positive operators.

For arbitrary square matrices M,N we write M ≥ N if M − N ≥ 0; i.e., 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 M > N.

1. Every positive definite matrix is invertible and its inverse is also positive definite. If M ≥ N > 0 then N⁻¹ ≥ M⁻¹ > 0.
2. If M is positive definite and r > 0 is a real number, then r M is positive definite. If M and N are positive definite, then the sum M + N and the products MNM and NMN are also positive definite. If MN = NM, then MN is also positive definite.
3. If M = (m_ij) > 0 then the diagonal entries m_ii are real and positive. As a consequence the trace, tr(M) > 0. Furthermore

   $|m_{ij}|\leq {\sqrt {m_{ii}m_{jj}}}\leq {\frac {1}{2}}(m_{ii}+m_{jj})$

   and thus

   $\max |m_{ij}|\leq \max |m_{ii}|$

4. A matrix M is positive semi-definite if and only if there is a positive semi-definite matrix B with B² = M. This matrix B is unique, is called the square root of M, and is denoted with B = M^1/2 (the square root B is not to be confused with the matrix L in the Cholesky factorization M = LL*, which is also sometimes called the square root of M). If M > N > 0 then M^1/2 > N^1/2 > 0.
5. If M,N > 0 then M ⊗ N > 0, where ⊗ denotes Kronecker product.
6. For matrices M = (m_ij) and N = (n_ij), write M · N for the entry-wise product of M and N; i.e., the matrix whose i,j entry is m_ijn_ij. Then M · N is the Hadamard product of M and N. The Hadamard product of two positive-definite matrices is again positive-definite and the Hadamard product of two positive-semidefinite matrices is again positive-semidefinite (this result is often called the Schur product theorem). Furthermore, if M and N are positive-semidefinite, then the following inequality, due to Oppenheim, holds:

   $\det(M\cdot N)\geq (\det N)\prod _{i}m_{ii}.$

7. Let M > 0 and N Hermitian. If MN + NM ≥ 0 (resp., MN + NM > 0) then N ≥ 0 (resp., N ≥ 0).
8. If M,N ≥ 0 are real matrices then $\scriptstyle \operatorname {tr} (MN)\,\geq \,0$.
9. If M > 0 is real, then there is a δ > 0 such that M > δI, where I is the identity matrix.

Block Matrices

A positive 2n × 2n matrix may also be defined by blocks:

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

Where each block is n × n. By applying the positivity condition, it immediately follows that A and D are hermitian, and C = B*.

We have that z^*Mz ≥ 0 for all complex z, and in particular for z = ( v, 0)^T. Then

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

A similar argument can be applied to D, and thus we conclude that both A and D must be positive matrices, as well.

Non-Hermitian matrices

A real matrix M may have the property that x^TMx > 0 for all nonzero real vectors x without being symmetric. The matrix

${\begin{bmatrix}1&1\\-1&1\end{bmatrix}}$

satisfies this property, because for all real vectors $x=(x_{1},x_{2})^{T}$ such that $x\neq 0$,

${\begin{bmatrix}x_{1}&x_{2}\end{bmatrix}}{\begin{bmatrix}1&1\\-1&1\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}=x_{1}^{2}+x_{2}^{2}>0.$

In general, we have x^TMx > 0 for all real nonzero vectors x if and only if the symmetric part, (M + M^T) / 2, is positive definite.

The situation for complex matrices may be different, depending on how one generalizes the inequality z^*Mz > 0 when considering M which aren't necessarily Hermitian. If z^*Mz is real for all complex vectors z, then the matrix M must be Hermitian. To see this, we define the Hermitian matrices A=(M+M^*)/2 and B=(M-M^*)/(2i) so that M=A+iB. Then, z^*Mz=z^*Az+iz^*Bz is real. By the Hermiticity of A and B, z^*Az and z^*Bz are individually real so z^*Bz must be zero for all z. So then B is the zero matrix and M=A, which is Hermitian.

So, if we require that z^*Mz be real and positive, then M is automatically Hermitian. On the other hand, we have that Re(z^*Mz) > 0 for all complex nonzero vectors z if and only if the Hermitian part, (M + M^*) / 2, is positive definite.

In summary, the distinguishing feature between the real and complex case is that, a bounded positive operator on a complex Hilbert space is necessarily Hermitian, or self adjoint. The general claim can be argued using the polarization identity. That is no longer true in the real case.

There is no agreement in the literature on the proper definition of positive-definite for non-Hermitian matrices.

See also

• Cholesky decomposition
• Positive-definite function
• Positive definite kernel
• Schur complement
• Square root of a matrix
• Sylvester's criterion
• Covariance matrix

References

• Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN 978-0-521-38632-6.
• Rajendra Bhatia. Positive definite matrices. Princeton Series in Applied Mathematics, 2007. ISBN 978-0691129181.