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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.

A complex Wishart matrix is the Gram matrix of an $n$-by-$m$ matrix of independent centered complex Gaussian random variables with an indicated variance convention. It is a positive semidefinite matrix. In the standard convention with joint entry density proportional to $e^{-\mathrm{Tr}{X}^{\ast }X}$ and $n\ge m$, its eigenvalue density contains the weight $x^{n-m}{e}^{-x}$ and a squared Vandermonde determinant.

A convex quadratic function on ${\mathbb{R}}^n$ has the form $q(x)=\frac{1}{2}{x}^{T}Hx+b^{T}x+c$ with a symmetric positive semidefinite matrix $H$. Its Hessian is $H$, and $q(y)-q(x)-\nabla q(x)\cdot (y-x)=\frac{1}{2}(y-x{)}^{T}H(y-x)\ge 0$, proving convexity directly. A stationary point is a global minimum. On a compact polytope, a minimum either is stationary in the interior or minimizes a restriction to a boundary face.

A real symmetric matrix $A$ is copositive if its quadratic form satisfies $x^{T}Ax\ge 0$ for every $x$ in the nonnegative orthant. Every positive semidefinite matrix and every symmetric nonnegative matrix is copositive, as is their sum. The converse fails for the Horn copositive matrix.

For $X\sim N(\mu 1,C)$ with a known positive-definite matrix $C$, the maximum-likelihood estimator of $\mu$ is $\hat{\mu }=(1^T{C}^{-1}X)/(1^T{C}^{-1}1)$. It is unbiased with variance $(1^T{C}^{-1}1{)}^{-1}$, attaining the Cramér-Rao bound. It is also the best linear unbiased estimator. Pairwise admissible correlation coefficients alone do not ensure that $C$ is a valid covariance matrix; the entire matrix must be symmetric and a positive semidefinite matrix. Inverse-based formulas require positive definiteness.

For a real random vector $X=(X_1,\dots ,X_p{)}^T$ with finite second moments, its covariance matrix collects all pairwise covariances:The diagonal entries are variances. The covariance matrix is a symmetric matrix and a positive semidefinite matrix, because for every real vector $a$,Thus one matrix describes the variance of every linear combination of the coordinates. It is a positive-definite matrix exactly when no nonzero linear combination is constant almost surely. A deterministic linear relation produces a zero eigenvalue.

For example,has unit coordinate variances, covariance $0.8$, and eigenvalues $1.8$ and $0.2$. Its unit eigenvectors $(1,1{)}^T/\sqrt{2}$ and $(1,-1{)}^T/\sqrt{2}$ describe directions of greatest and least projected variance. For a unit vector $a=(\cos \theta ,\sin \theta {)}^T$, direct multiplication gives $a^T\Sigma a=1+0.8\sin (2\theta )$: the variation in the sum direction is nine times that in the difference direction.

The spectral theorem supplies an orthonormal basis of these directions. This connects covariance matrices to principal component analysis: its leading eigenvector identifies the unit-direction projection with the largest variance. For a linear transformation $A$, the transformed random vector has covariance matrix $A\Sigma A^T$, so uncertainties propagate through a linear model by matrix multiplication.

If the weight vector lies in the linear span of the training feature map outputs, write $w={\sum }_i{\beta }_i\varphi (p_i)$. ThenThis is a dual representation: coefficients weight training examples rather than feature coordinates. They need not be positive, and need not be unique when the kernel matrix is singular. Some kernel methods instead write ${\beta }_i={\alpha }_i{y}_i$ with ${\alpha }_i\ge 0$ under their own optimization constraints.^[ref] The formula is an algebraic conversion for a weight vector in this linear span, not a claim about every possible weight vector. The representer theorem gives a broader reason this form appears in regularized kernel methods.

For the XOR separation by a product feature, let $\Phi$ have rows $\varphi (p_i{)}^T$. The Gram matrix of those rows, equivalently the kernel matrix, isIts eigenvalues are $8,8,0,0$, so it is a positive semidefinite matrix of rank two, not an invertible matrix. More generally, for any real coefficient vector $a$ and any finite set of inputs,This proves the positive-semidefinite kernel condition for the displayed polynomial kernel on all of ${\mathbb{R}}^2$, rather than checking only this square.

Here $Ky=8y$, so the explicit choice $\beta =y/8$ gives $K\beta =y$ and ${\Phi }^T\beta =w$. ConsequentlyFor the new input $z=(1,1/2)$, the four polynomial kernel evaluations are $(9/4,1/4,1/4,9/4{)}^T$. Their weighted sum is $-1/2$, and the sign function predicts $-1$. No explicit feature map is needed for this calculation. These coefficients were constructed to reproduce the chosen separator; no fitting algorithm or generalization guarantee is being asserted.

For a real positive-definite matrix $A$ in finite dimension, write $A=Q\Lambda Q^T$ by the spectral theorem for real symmetric matrices. In the coordinates in an orthonormal basis $w=Q^{T}x$, the positive level set $q(x)=k$ becomesThis is an ellipsoid, with semi-major and semi-minor axes in two dimensions forming an ellipse; each semi-axis has length $r_j=\sqrt{k/{\lambda }_j}$ and direction given by the corresponding eigenvector. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_{+}$, while the longer is $\sqrt{k/2}$ along $e_{-}$. A stiffer direction, meaning a larger eigenvalue, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{{\lambda }_{\max }/{\lambda }_{\min }}$, the square root of the spectral condition number of a positive-definite matrix.^[ref]

Strict positivity matters because ${\lambda }_{\min }>0$ givesHence a fixed positive level set cannot extend arbitrarily far. The Cholesky decomposition $A=L{L}^T$ gives another view: $q(x)=\Vert L^{T}x{\Vert }^2$, so the ellipsoid is the inverse image of a sphere under an invertible linear map. A singular positive semidefinite matrix has a nonzero kernel; adding a kernel vector preserves its quadratic form, so any nonempty positive level set is unbounded. These statements concern finite dimension; the positive definite symmetric operator entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional Hilbert space.

The set of real positive semidefinite matrices with unit diagonal. Equivalently, its members are Gram matrices of unit vectors, by the real spectral theorem. Every two-by-two principal minor gives $|X_{ij}|\le 1$, so the set is bounded; it is also closed, hence compact in finite dimension. This gives attainment of linear objectives used in semidefinite programming.

A nonzero Hermitian rank-one positive semidefinite matrix has zero determinant and positive trace, representing a future-directed null vector. It factors as $\chi {\chi }^{†}$, with an irrelevant spinor phase. A past-directed null vector instead requires a minus sign; a positive outer product alone does not cover both time directions.

With canonically normalized gauge field kinetic terms, a gauge-boson mass matrix is the real symmetric matrix of coefficients in the displayed quadratic mass term. Its eigenvalues are squared physical gauge boson masses. For a complex scalar field vacuum ${\Phi }_0$ in the Higgs mechanism, $(M^2{)}_{ab}=2\mathrm{Re}[(g_a{T}_a{\Phi }_0{)}^{†}(g_b{T}_b{\Phi }_0)]$. Thus for real $x_a$, $x^T{M}^{2}x=2\Vert {\sum }_a{x}_a{g}_a{T}_a{\Phi }_0{\Vert }^2\ge 0$: it is a Gram matrix and a positive semidefinite matrix. An unbroken generator annihilates the vacuum expectation value and lies in its null space. Diagonalizing the matrix identifies the massive and massless field combinations.

Observe $Y_t=\lambda t+W_t$, with an independent Gaussian random vector $\lambda$ of mean $m_0$ and covariance matrix $V_0$. Its posterior is Gaussian with mean $m_t=(I+t{V}_0{)}^{-1}(m_0+V_0{Y}_t)$ and covariance matrix $V_t=V_0(I+t{V}_0{)}^{-1}$. These formulas also hold for a singular positive semidefinite matrix $V_0$. When $V_0$ is invertible, completing the square in the Bayes' theorem gives $V_t^{-1}=V_0^{-1}+tI$ and $m_t=V_t(V_0^{-1}{m}_0+Y_t)$. The observed innovation process ${\hat{W}}_t=Y_t-{\int }_0^t{m}_{s}ds$ is a Brownian motion in the observation filtration, and $d{m}_t=V_{t}d{\hat{W}}_t$.

A real matrix product giving pairwise dot products of columns, for $T^{\top }T$, or rows, for $T{T}^{\top }$. It is a positive semidefinite matrix. The kernel of a Gram product is determined by the corresponding factor, which makes these products useful for converting a zero quadratic form into an exact null space condition.

Assign a real value $x_i$ to each vertex. The Graph Laplacian quadratic form measures disagreement across edges:This makes the Graph Laplacian a positive semidefinite matrix. The energy is small when neighbours have similar values, even if distant vertices have very different values; it depends on connections rather than where the graph is drawn. In the three-vertex path graph, take $x=(2,0,-1{)}^T$. Thenand the two edges contribute $(2-0{)}^2=4$ and $(0-(-1){)}^2=1$.

For a short proof, expand $x^T(D-A)x$. Each edge contributes $x_i^2+x_j^2$ through the degree matrix and $-2x_i{x}_j$ through the adjacency matrix, yielding $(x_i-x_j{)}^2$. The sum over unordered edges has no factor $1/2$; a sum over all ordered pairs does:Equivalently, let $B$ be an oriented incidence matrix, with one row per edge. Then $Bx$ records endpoint differences and $L=B^{T}B$, independently of the chosen orientations. This translates a local comparison into a global quadratic form and makes its nonnegativity visible.

For integer $k\ge 1$, the entrywise power has entries $X_{ij}^k$ and equals the repeated Hadamard product. The zero power is the all-ones matrix, even at zero entries, because it represents the constant function one. Every nonnegative integer Hadamard power of a positive semidefinite matrix is positive semidefinite by the Schur product theorem. It is not an ordinary matrix power.

The matrix with entries $(P\circ Q{)}_{ij}=P_{ij}{Q}_{ij}$. It differs from ordinary matrix multiplication. The Schur product theorem states that it preserves positive semidefiniteness when both factors are positive semidefinite matrices.

On an open convex set, a twice-differentiable real-valued function is convex if and only if its Hessian matrix is a positive semidefinite matrix at every point. In particular continuous second partial derivatives suffice for the differentiability hypothesis. Restrict $g$ to any line segment: $\varphi (t)=g(x+tv)$ has ${\varphi }^{\prime \prime }(t)=v^{\mathsf{T}}H(g)(x+tv)v$. The second-derivative criterion for convexity proves both implications, because convexity is equivalent to convexity of all such restrictions.

The five-dimensional Horn copositive matrix has diagonal entries $1$, entries $-1$ on the edges of the five-cycle, and entries $1$ on the remaining pairs:For $x\ge 0$, a cyclic relabelling puts a smallest coordinate at $x_5$. The identitythen proves that $H$ is a copositive matrix.

However, $H$ is outside the positive-semidefinite-plus-nonnegative cone. Set $w=(1,2,1,0,0{)}^T$. Its quadratic form is zero. If $H=P+N$ with $P$ a positive semidefinite matrix and $N$ a symmetric nonnegative matrix, both $w^{T}Pw$ and $w^{T}Nw$ must vanish. Positivity of the first three coordinates of $w$ forces every entry of the leading $3\times 3$ block of $N$ to vanish. Applying the argument to all cyclic shifts of $w$ forces every entry of $N$ to vanish, since each pair of indices lies in a cyclic interval of length three. This would make $H$ a positive semidefinite matrix, but zero quadratic form of a positive semidefinite matrix would then give $Hw=0$, whereas $Hw=(0,0,0,2,2{)}^T$.

The Kalman measurement update incorporates the new observation by Gaussian conditional expectation:The posterior covariance matrix subtracts a positive semidefinite matrix, reflecting the information supplied by the measurement. Even if the innovation is zero and the mean does not move, the uncertainty can decrease. In this linear Gaussian model the posterior covariance matrix depends on the observation geometry and its uncertainty, rather than on the numerical observed value.

The derivation is one application of Gaussian conditional expectation per measurement. Before observing $y_t$, the state and observation have joint covariance matrixConditioning the upper block on the lower block gives exactly the two update equations. The upper-right block explains the Kalman gain: a measured component corrects a state component only to the extent that their prediction errors have covariance. The measurement can therefore improve an unobserved velocity when a state-space model correlates its error with an observed position.