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On positive-determinant deformation gradients, the first Piola-Kirchhoff stress is $\mu (F-J^{-1}{F}^{-T})$. Its zero set is $SO(3)$: multiplication by $F^T$ gives $F{F}^T=J^{-1}I$, and determinants give $J^5=1$. These are also the global minimizers, since $|F{|}^2\ge 3J^{2/3}$ and $3J^{2/3}/2+J^{-1}$ has its unique minimum $5/2$ at $J=1$. The minimum energy is $5\mu /2$ rather than zero; an additive constant changes no stress.

The reduced norm of an element of a central simple algebra of degree $d$ is its determinant after any splitting isomorphism to $M_d$. Inner automorphisms of a matrix algebra preserve that determinant, while Galois invariance descends its polynomial coefficients to the center. It is a multiplicative homogeneous degree-$d$ polynomial, vanishing exactly on noninvertible elements. On a central division algebra it is anisotropic.

A Hartree-Fock ansatz in which opposite-spin electrons in a doubly occupied orbital share the same spatial function. The closed-shell determinant is spin-pure but can dissociate incorrectly when a bond requires several determinants.

The resultant turns a shared-root of a polynomial question into a calculation using only coefficients. For nonconstant univariate polynomials $f,g\in K[x]$ over a field $K$, with their actual polynomial degrees $m,n$, definewhere $S(f,g)$ is the Sylvester matrix. ThenHere $\overline{K}$ is an algebraic closure. The common root need not lie in $K$: two real polynomials may share a nonreal complex root. The subscript identifies which variable is eliminated.^[ref]

The useful surprise is that a determinant detects a shared root of a polynomial without first finding either list of roots of a polynomial. If the coefficients depend on a parameter, the resultant itself is a polynomial in that parameter. Its zeros pick out possible exceptional parameter values. Elimination theory uses this to reduce a pair of polynomial equations in two variables to a single necessary equation in one variable; eliminating a variable with a resultant gives an exact example below.

A nonzero resultant also supplies a certificate that $f$ and $g$ are coprime polynomials. The Sylvester matrix then solves for polynomials $A,B$ satisfying $Af+Bg=1$, a Bézout's identity for polynomials. Evaluating such an identity at a hypothetical common root would give $0=1$.

For a compact Riemann surface of genus $g\ge 1$, the theta divisor in ${\mathrm{Pic}}^{g-1}(X)$ has the displayed multiplicity. A local determinant description of the theta divisor reduces to an $r$ by $r$ matrix vanishing at $L$, where $r=h^0(L)$. Its linear term is cup product with a tangent class in $H^1({\mathcal{O}}_X)$. The invertible cup-product direction for a line bundle on a curve shows that its determinant has order exactly $r$. Consequently the singular locus consists precisely of classes with at least two independent global sections.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

The Robertson-Schrödinger uncertainty relation keeps the real part discarded in the proof of the Robertson uncertainty principle. Write $A^{\prime }=A-⟨A⟩I$ and $B^{\prime }=B-⟨B⟩I$. Their quantum covariance is $C_{AB}=\mathrm{Re}⟨A^{\prime }\psi |B^{\prime }\psi ⟩$. On the same finite-moment and product operator domains used above, the Cauchy-Schwarz inequality givesFor position and momentum, form the real, symmetrized covariance matrixA larger coordinatewise product can therefore encode a tilted covariance ellipse, rather than a larger determinant. This explains why the equality case of the Heisenberg uncertainty relation is more restrictive than equality in the Robertson-Schrödinger uncertainty relation.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

The determinant transformation formula for the Levi-Civita symbol shows that it is invariant under every proper rotation matrix, for which $\det R=1$. Under an orthogonal reflection its sign reverses. This distinguishes proper-rotation isotropy from the pseudotensor transformation convention when both orientations are included.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]

The Routh-Hurwitz criterion tests whether all roots of a real polynomial have negative real part, by sign conditions on determinants formed from its coefficients. Applied to a characteristic polynomial, it gives an algebraic test for strict linear asymptotic stability.

For a finite-index subgroup $L\le {\mathbb{Z}}^2$, its projection to the first coordinate is $a\mathbb{Z}$ and its intersection with the second axis is $\{0\}\times d\mathbb{Z}$, with $a,d>0$. Choose $(a,b)\in L$, reducing $b$ modulo $d$. Together with $(0,d)$ it is a basis of this row lattice: subtracting a multiple of $(a,b)$ from any vector leaves a vector on the second axis. These parameters are unique, and reduction of the two coordinates shows $[{\mathbb{Z}}^2:L]=ad$. Consequently an integral matrix of positive determinant $n$ has a unique representative of this form under left multiplication by $S{L}_2(\mathbb{Z})$, with $ad=n$. This proves the lattice facts underlying determinant-n matrix representatives for Hecke operators.

Exchanging two rows of a matrix. It is its own inverse and changes the sign of the determinant of a square matrix.

The row lattice of an integer matrix with $m$ rows is the set of their integer linear combinations. It is an Euclidean lattice in its real span, since it is contained in the discrete set ${\mathbb{Z}}^n$ and spans that real vector space. For a nonsingular two-by-two integer matrix, the row Hermite normal form in rank two gives a canonical basis and proves that its index in ${\mathbb{Z}}^2$ is the absolute value of its determinant.

Work over the rational numbers, allowing unknowns in the real numbers when interpreting the geometry. Consider four equations in three unknowns:The third row of $A$ is the sum of its first two rows, and the fourth is their difference, second minus first. The same relations must hold on the right: consistency requires $\tau =1+4=5$ and $3=4-1$. The latter condition already holds. Gaussian elimination makes this compatibility test explicit while exposing the remaining freedom.

Apply the elementary row operations in this order, using each row's current value:then $R_3\leftarrow R_3-R_2$ and $R_4\leftarrow R_4-R_2$. The resulting augmented matrix is in row echelon form for every $\tau$:When $\tau \ne 5$, the third row has its pivot element in the augmented column. It says $0=\tau -5$, so the linear system has no solution. When $\tau =5$, that entire row vanishes and only two pivot elements remain. More equations than unknowns therefore do not force a unique solution: some equations may repeat information.

The final operation $R_1\leftarrow R_1-2R_2$ reduces the coefficient block toThe product of all six elementary matrices isDirect matrix multiplication verifies both displayed transformations. Its determinant is one, so this transformation is reversible. In particular, the third row of $E$ extracts the discrepancy between the third equation and the sum of the first two.

For $\tau =5$, $[R\mid Eb]$ is the reduced row echelon form of the full augmented matrix. For $\tau \ne 5$, only its coefficient block is reduced: scale the third row by $1/(\tau -5)$ and clear the entries above its augmented pivot element. The full reduced row echelon form is thenThe last column is part of the matrix when testing the echelon conditions. Calling $[R\mid Eb]$ fully reduced before clearing that column would miss this distinction.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

When $A$ is an invertible matrix, the factorization in block elimination has outer triangular factors with determinant one. Hence $\det M=\det A\det S$, without a symmetry assumption. For the original example under block matrix inverse, this gives $\det M=(2\cdot 3)(19/6)=19$. An invertible block $A$ therefore reduces the question of whether the full matrix is an invertible matrix to the smaller Schur complement; a singular $S$ makes $M$ a singular matrix.

For a square evaluation matrix representing a degree-$g-1$ line bundle $L$, remove an invertible minor of size $N-h^0(L)$. Holomorphic row and column operations leave a square matrix $M$ of size $r=h^0(L)$ with zero constant term. Its linear part in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is $\xi ⌣-:H^0(L)\to H^1(L)$. The invertible cup-product direction for a line bundle on a curve makes the leading degree-$r$ determinant nonzero, establishing order exactly $r$ rather than just a lower bound.

Put $b_i={\lambda }_i+N-i$. In the bialternant formula, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ and compare the first nonzero determinant coefficients as $t\to 0$. They arise from the distinct degrees $0,\dots ,N-1$ in the exponential expansions. The common factor in the $y_j$ and the factorials cancel, leaving the product of differences of the $b_i$ divided by the differences of $N-i$. This proves the displayed integral specialization without dividing by a vanishing Vandermonde value.

At normal incidence, a uniform lossless elastic wave layer of thickness $h_j$, speed $c_j$ and seismic impedance $q_j={\rho }_j{c}_j$ has wave phase thickness ${\delta }_j=\omega h_j/c_j$. With phasors proportional to $e^{-i\omega t}$, particle velocity $V$ and compressive traction $P=-T$, the displayed matrix maps $(V,P)$ from the top to the bottom. It has determinant one. Multiplying these matrices in physical order accounts for all coherent multiple reflections. Continuity of $V,P$ connects adjacent layers without another propagation factor.