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An SU(2) matrix is a two-by-two complex unitary matrix with determinant one. It has the form $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Its action on the Riemann sphere is a rotational Möbius transformation; opposite matrices induce the same sphere rotation.

On ${\mathbb{C}}^3\oplus \mathbb{R}$, take $\omega =(i/2){\sum }_{j}d{z}_j\wedge d{\bar{z}}_j$ and $\Omega =d{z}_1\wedge d{z}_2\wedge d{z}_3$. The special unitary group $SU(3)$ preserves the Kähler form $\omega$ by unitarity and preserves $\Omega$ by its unit determinant. Fixing $y$ therefore embeds it in the compact G2 stabilizer of $\varphi$. The complexified seven-dimensional representation restricts as $3\oplus \bar{3}\oplus 1$.

The four-by-four unitary matrices of determinant one form this special unitary group. Its two-index antisymmetric representation has complex dimension six and the real structure of the SU4 exterior square realizes it on a six-dimensional real space.

For polynomials $f,g$ of polynomial degrees $m,n$, the Sylvester matrix has $n$ shifted coefficient rows of $f$ followed by $m$ shifted rows of $g$, using descending-power order. It is a square matrix of size $m+n$. Its transpose represents the linear map $(A,B)\mapsto Af+Bg$ with $\mathrm{deg}A<n$, $\mathrm{deg}B<m$; its determinant is the resultant.

For an original parameter example, choose $f(x)=x^2-1$ and $g_t(x)=x-t$. ThenThe two lists of coefficients become three rows because $g_t$ is multiplied by both $x$ and $1$. Thus the common-root parameter values are $t=-1$ and $t=1$. At $t=1/2$, the resultant is $-3/4$, so the two polynomials have no common root.

One can verify the certificate without any determinant expansion:For $t\ne \pm 1$, division by $t^2-1$ gives a Bézout's identity for polynomials. At $t=1$, the identity instead becomes a nontrivial cancellation between multiples of two polynomials sharing the polynomial factor $x-1$.

The general proof needs only a short linear algebra outline. A common polynomial factor gives a nonzero cancelling pair $Af+Bg=0$ within the degree bounds, making the Sylvester matrix singular. Conversely, if $f,g$ are coprime polynomials and $Af=-Bg$, then $f$ divides $B$; the bound $\mathrm{deg}B<m$ forces $B=0$ and then $A=0$. Thus the linear map has zero kernel of a linear map exactly in the coprime case. Over an algebraic closure, a nonconstant common polynomial factor is equivalent to a common root.

In 1840, James Joseph Sylvester described forming the shifted coefficient rows and taking a signed sum of products selected by permutations. In modern notation, this is the Sylvester matrix and its determinant. The construction turns the nonlinear common-root of a polynomial question into linear algebra. His paper's word derivatives concerns elimination, rather than the formal derivative used below.^[ref]

A Hermitian matrix $A$ is a positive-definite matrix exactly when its leading principal minorsare all positive. The leading principal minor uses the first $k$ rows and the first $k$ columns in the same ordering. For the two-by-two Cholesky decomposition example, ${\Delta }_1=4$ and ${\Delta }_2=4\cdot 5-2^2=16$, so the criterion proves positive definiteness without calculating eigenvalues.

The successive pivot elements before taking square roots are ${\Delta }_k/{\Delta }_{k-1}$, with ${\Delta }_0=1$. Here they are $4$ and $16/4=4$. In general, a leading triangular matrix factor has determinant equal to the product of its diagonal entries, soConversely, positive leading principal minors make each successive Schur complement pivot element positive, allowing the Cholesky decomposition to be constructed. The identity $u^{T}Au=\Vert L^{T}u{\Vert }_2^2$ then proves positive definiteness; for complex matrices, replace the transposes by conjugate transposes. This gives a short inductive proof of the criterion.

For $B=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, the first minor is positive but $\det B=-3$. The second Cholesky radicand would be $1-2^2=-3$, and $v=(1,-1{)}^T$ directly gives $v^{T}Bv=-2<0$. Thus positive diagonal entries are insufficient. Replacing “positive” by “nonnegative” in just the leading-minor test is also insufficient for positive semidefiniteness: $\mathrm{diag}(0,-1)$ has leading minors $0,0$, but its quadratic form is negative on $(0,1{)}^T$.

Two identical concave mirrors of curvature radius R separated by L have half-round-trip ray matrix $A=\left(\begin{array}{cc}1& L\\ -2/R& 1-2L/R\end{array}\right)$. Its determinant is one and its trace is $2(1-L/R)$. The eigenvalues are a conjugate unit-modulus pair for the displayed interval. At the confocal point $L=R$, $A^2=-I$, so it is bounded despite the full-round-trip trace reaching minus two. Matching a Gaussian beam's mirror wavefront gives $z_R^2=(L/2)(R-L/2)$ and $\pi w_0^2=\lambda z_R$. The endpoints are degenerate limiting geometries, not finite confined Gaussian modes.

A real symmetric tensor of the special orthogonal group transforms as $Q\mapsto OQ{O}^T$. If its vacuum expectation value has distinct eigenvalues with multiplicities $m_1,\dots ,m_p$, its stabilizer subgroup is $S(O(m_1)\times \cdots \times O(m_p))$: each transformation preserves each eigenspace and the product of the block determinants is one. Its identity component is ${\prod }_{i}SO(m_i)$. Thus a scalar multiple of the identity preserves the whole group, whereas a tensor with simple spectrum preserves only diagonal signs with total sign one. A tracelessness condition constrains the eigenvalues but does not change this stabilizer formula.

The tangent cone to a theta divisor at a degree-$g-1$ line bundle $L$ is cut out in $H^1({\mathcal{O}}_X)$ by the determinant of $\xi ⌣-:H^0(L)\to H^1(L)$. Under Serre duality its matrix entries are $⟨\xi ,s_i{t}_j⟩$ for bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$. For $h^0(L)=1$, its tangent hyperplane is annihilated by the nonzero holomorphic differential form $s_1{t}_1$.

Take a central chemical atom at the origin and four substituents at the vertices of an ideal regular tetrahedron:These dimensionless coordinates describe shape, not measured bond lengths. For distinct labels A, B, C and D, define a signed determinantIt is six times the oriented volume of the labelled tetrahedron. A translation cancels from the differences. A rotation $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror matrix $S=\mathrm{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no rotation can place every distinct label on its mirror partner. This argument uses three-dimensional coordinates, not clockwise order on a page.

Now let C and D both be the same substituent H. Swapping these identical groups is allowed. For this model the half-turnabout the axis $(0,1,1)$ maps the reflected A and B to their original positions and exchanges the two H positions. The models superpose. With distinct C and D, the same half-turn fails precisely at those two labels; the determinant argument rules out every other rotation as well. This is the local reason two identical hydrogen chemical atoms prevent glycine's amino acid alpha carbon from being a tetrahedral stereocenter.

For a flexible molecule, also distinguish conformational isomerism from a persistent enantiomer pair. A particular frozen shape may be chiral while ordinary internal motions interconvert mirror shapes. Turning the entire model is always allowed in a spatial comparison; an internal motion must be considered with its actual accessibility and timescale.

For cube vertices $(\pm 1,\pm 1,\pm 1)$, the four vertices of coordinate product $+1$ and the four of product $-1$ form the two inscribed regular tetrahedra. A cube rotation is a signed permutation matrix of determinant one and maps each product class to one class. A quarter-turn about a coordinate axis exchanges the classes. The orbit of either tetrahedron therefore has size two, while the cube rotation group has order 24. The orbit-stabilizer theorem gives its setwise stabilizer subgroup order 12; this is the tetrahedron's proper rotation group.

For a number-conserving Gaussian pairing state, six permutations give this expression. Each transposition has sign $ϵ$, positive for bosons and negative for fermions; the two three-cycles have positive sign for either statistics. Equivalently it is the permanent or determinant of the three-by-three one-body density matrix. For bosonic number states, repeated-mode factorial corrections must be included instead of assuming that Wick theorem.

Evaluation determinants for $L\otimes P$ and $L$ have a quotient which is a rational section of ${⨂}_i{\mathrm{pr}}_i^{\ast }P$, not intrinsically a scalar. If $P^N\cong {\mathcal{O}}_C$, its $N$th power becomes a rational function after choosing that trivialization. In Cartier divisor notation, write $P={\mathcal{O}}_C(p)$ and $\mathrm{div}(h)=Np$. With raw meromorphic evaluations the scalar expression is $r^N{\prod }_{i}h(z_i)$. The extra product accounts for the line-bundle frames; it is suppressed when evaluation already uses compatible torsion trivializations.

The directed vertex-edge incidence matrix has one $+1$ and one $-1$ per nonloop column and a zero column for a loop. Every square subdeterminant belongs to $\{0,1,-1\}$. Induct on its size: a column with no nonzero entries makes the determinant zero; a column with one nonzero entry permits a cofactor expansion to a smaller determinant; and if every column has two nonzero entries, their opposite signs make the column sums zero, giving linearly dependent rows. Negating rows preserves absolute subdeterminants. Consequently the unsigned vertex-edge incidence matrix of a bipartite graph is totally unimodular: negate all rows in one part. The assignment problem is an instance, with row vertices and column vertices forming the two parts.

For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.

For $P=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the off-diagonal equations in $P^2=P$ are $b(a+d-1)=c(a+d-1)=0$. Thus either $b=c=0$ or the trace is one. The condition alone is not sufficient: $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has trace one but is not an idempotent matrix. For a two-by-two matrix of trace one, idempotence is equivalent to zero determinant, by the Cayley-Hamilton theorem.

For a finite-dimensional real or complex linear projection $P$, every vector splits as $Px+(I-P)x$ in the image of a linear map and the kernel of a linear map, respectively. Their intersection is zero, so a basis adapted to this direct sum represents $P$ as $\mathrm{diag}(I_r,0)$. Its trace and rank both equal $r$. In particular, a projection's determinant is zero unless it is the identity. Hermiticity is unnecessary for this algebraic identity; a Hermitian matrix projection is additionally orthogonal.

For a real-frequency lossless reciprocal periodic line, a cell transfer matrix has determinant one. If half its trace lies outside $[-1,1]$, the Bloch multipliers are real reciprocal numbers rather than unit-modulus propagation factors. The physical interior field decays and carries no net power to infinity. A real reference line is then totally reflected by a semi-infinite structure; a finite structure can still tunnel.

The trefoil is the torus knot $T_{2,3}$. It is the simplest nontrivial knot by crossing number of a knot: $c(T_{2,3})=3$. A standard three-crossing alternating knot diagram is reduced, so the Tait crossing-number theorem establishes minimality. The four-crossing knot diagram shown below also represents the trefoil knot, illustrating that a particular projection need not achieve that minimum. Its Seifert genus is one.

Here is a small algebraic certificate that a trefoil knot cannot be deformed into an unknot. One Seifert matrix for a trefoil knot isTaking the determinant gives the Alexander polynomial of a knotwhere $\doteq$ allows multiplication by a Laurent unit $\pm t^n$ in the Laurent polynomial ring. No such multiplication turns this three-term Alexander polynomial of a knot into the unknot's value $1$. Since the Alexander polynomial of a knot is a knot invariant, this computation rules out every possible ambient isotopy between the two knots; it does not merely rule out the deformations we happened to try.

The knot group supplies another certificate. The presentation $⟨a,b\mid a^2=b^3⟩$ maps onto the symmetric group $S_3$ by $a\mapsto (12)$ and $b\mapsto (123)$, since both sides of the relation map to the identity. These images generate a non-abelian group, so the knot group cannot be the unknot's infinite cyclic group. However, a trefoil knot and its mirror of a link have isomorphic knot groups: the abstract knot group alone does not distinguish all knots.

Add diffusivities $1,p$ to the basally forced quadratic activator-inhibitor model. A Fourier mode with $s=k^2$ has matrix $J-s\mathrm{diag}(1,p)$ and determinant $D(s)=p{s}^2+(q-pa)s+q$. When $q>a$, its trace is negative for every $s\ge 0$. A positive-growth mode therefore occurs precisely when $D(s)<0$ for some positive $s$. The parabola has a positive minimizer only if $pa>q$, and its minimum is negative exactly when $(pa-q{)}^2>4pq$. Thus the ordinary Turing instability requires $a>0$, $q>a$ and $p>q/(\sqrt{1+a}-1{)}^2$. At onset, $k_c^2=(pa-q)/(2p)=\sqrt{q/p}$ and the critical wavelength is $2\pi (p/q{)}^{1/4}$. A bounded spatial domain further requires an allowed mode inside the unstable band.

In cylindrical orthonormal bases the deformation gradient is $F=\left(\begin{array}{ccc}\rho /r& 0& 0\\ 0& r/\rho & \alpha r\\ 0& 0& 1\end{array}\right)$ and has unit determinant. One principal stretch is $\rho /r$; the squares of the other two are the roots of $z^2-[(r/\rho {)}^2+{\alpha }^2{r}^2+1]z+(r/\rho {)}^2=0$. The reference-volume energy per height is $E=2\pi \int \rho Wd\rho$. Virtual work gives end torque $M={\partial }_{\alpha }E$ and inner pressure $p=(2\pi a{)}^{-1}{\partial }_{a}E$ at fixed reference radii and height.