Positive semidefinite matrix

A real symmetric or complex Hermitian matrix $A$ is positive semidefinite when $x^{†}Ax\ge 0$ for every vector $x$, equivalently when all its eigenvalues are nonnegative.

For positive block sizes, the real matrix $\left(\begin{array}{cc}aJ& bJ\\ bJ& aJ\end{array}\right)$ is a positive semidefinite matrix exactly when $a\ge |b|$. For test-vector block sums $s,t$, its quadratic form is $(a+b)(s+t{)}^2/2+(a-b)(s-t{)}^2/2$. This proves sufficiency; testing $s=\pm t$ proves necessity. Zero sums in both blocks describe part of its kernel.

A real symmetric matrix admitting a finite sum of nonnegative rank-one matrices $x_j{x}_j^T$. It is both a positive semidefinite matrix and a nonnegative matrix, but these two properties alone need not imply membership in the completely positive cone in arbitrary dimension. This matrix notion is distinct from a completely positive map.

If $P$ is a real positive semidefinite matrix, then $x^{T}Px=0$ if and only if $Px=0$. Indeed $x^{T}Px=\Vert \sqrt{P}x{\Vert }^2$ by the principal square root of a positive semidefinite matrix, so a zero quadratic form puts $x$ in the kernel of $\sqrt{P}$ and hence of $P$.

For real positive semidefinite matrices $A,B$, their product need not be symmetric, but its matrix trace is nonnegative:The equality uses the cyclic property of the matrix trace, and the last matrix is a positive semidefinite matrix. Consequently the Loewner order inequality $A⪯C$ implies $\mathrm{tr}(AB)\le \mathrm{tr}(CB)$ whenever $B⪰0$.

The Loewner order on Hermitian matrices is defined by $A⪯B$ when $B-A$ is a positive semidefinite matrix.

A square root of a square matrix $A$ is a matrix $B$ satisfying $B^2=A$.

If $A=Q\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)Q^T$ is real symmetric and positive semidefinite, its principal square root isIt is symmetric and positive semidefinite, and is positive definite exactly when $A$ is.

Every invertible real matrix has the polar decompositionwhere $P$ is symmetric positive definite and $R$ is orthogonal.

Every invertible complex matrix has the unique polar decompositionwhere $P$ is a positive-definite Hermitian matrix and $U$ is a unitary matrix. Writing $P=e^B$ gives $B=\frac{1}{2}\log (M{M}^{\ast })$, which is Hermitian.