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The acceleration tidal tensor is $T_{jk}={\partial }_k{g}_j=-{\partial }_j{\partial }_k\Phi$ for gravitational acceleration $\mathbf{g}=-\nabla \Phi$. It describes differential acceleration $\Delta \mathbf{g}\simeq T\Delta \mathbf{r}$. Some authors instead call the positive Hessian matrix ${\partial }_j{\partial }_k\Phi$ the tidal tensor, reversing this sign convention. For a point mass and the positive potential convention $\mathbf{g}=\nabla \psi$, the tensor is $Gm(3\mathbf{R}{\mathbf{R}}^{\mathsf{T}}/R^5-1/R^3)$.

The tidal tensor is the Jacobian matrix of the Newtonian gravitational field, rather than the gravitational acceleration itself. Its entries have units ${\mathrm{s}}^{-2}$: multiplying by a displacement produces an acceleration. For a smooth Newtonian gravitational potential, the tidal tensor is a symmetric matrix, since mixed partial derivatives commute. A spatially uniform gravitational acceleration has zero tidal tensor, however large that acceleration is.

For real $a,b$, the displayed symmetric matrix satisfies $M^2=(a^2+b^2)I$. If $s=\sqrt{a^2+b^2}>0$, choose $\theta$ with $\cos 2\theta =a/s$ and $\sin 2\theta =b/s$. The eigenvectors $(\cos \theta ,\sin \theta )$ and $(-\sin \theta ,\cos \theta )$ are orthonormal and have eigenvalues $s$ and $-s$. The half-angle description works even when $b=0$, avoiding an artificial division by an off-diagonal entry. If $s=0$, the matrix is zero and every direction is an eigenvector. The normalized matrix $M/s$ is reflection in the axis at angle $\theta$.

Let $S$ be a real symmetric matrix and $A$ a real antisymmetric matrix of the same size. Invariance of the trace under transpose and the cyclic trace identity giveHence $\mathrm{tr}(SA)=0$. Equivalently these two matrices are orthogonal for the Frobenius inner product, since $⟨S,A{⟩}_F=\mathrm{tr}(S^{T}A)=\mathrm{tr}(SA)$. Together with the unique symmetric and antisymmetric parts of a matrix, this is an orthogonal decomposition of the real square matrices.

For a nearest-neighbour classical chain with finitely many spin states and dimensionless bond energy $E(s,t)$, its transfer matrix is $W_{st}=e^{-E(s,t)}$. On-site energies are divided between adjacent bonds. Periodic boundary conditions give partition function $Z_N=\mathrm{tr}{W}^N$ by direct multiplication and summation over spin labels. For a strictly positive symmetric matrix, the Perron–Frobenius theorem gives ${\lim }_{N\to \infty }{N}^{-1}\log Z_N=\log {\lambda }_{\max }$. A diagonal local observable $O$ has expectation $\mathrm{tr}(O{W}^N)/Z_N$, tending to $v_{\max }^{T}O{v}_{\max }$ for a normalized dominant eigenvector.

A trust region restricts a local approximation to a neighborhood where it is expected to model the objective function accurately. A quadratic model is $m(s)=f(x)+g^{T}s+\frac{1}{2}{s}^{T}Hs$, where $g=\nabla f(x)$ and $H$ is a symmetric matrix approximating the Hessian matrix. Minimizing this quadratic function over a Euclidean ball of radius $\Delta >0$ is the trust-region subproblem. The trust-region acceptance ratio then checks the model against the actual objective function.

For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.

Let a real symmetric matrix have an orthonormal eigenbasis $w_1,\dots ,w_n$, withWrite two starting vectors asFor two-column subspace iteration to recover
$\mathrm{span}(w_{n-1},w_n)$, the two projections onto that dominant subspace must be independent. The exact condition isRequiring each of the four coefficients to be nonzero does not imply this determinant condition.

A real Wigner matrix is a real symmetric matrix whose upper off-diagonal entries are independent centered copies of a fixed random variable, scaled to have variance of order $1/N$. Diagonal conventions may vary and should be specified separately. A common version has variance-one unscaled off-diagonal entries and either zero diagonal entries or independent diagonal entries with a bounded second moment.