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For the unitary group, the diagonal maximal torus has coordinates $z_i=e^{i{\theta }_i}$ and Weyl group $S_n$. The Weyl integration formula weights independent uniform angles by $n{!}^{-1}{\prod }_{i<j}|z_i-z_j{|}^2$. The Vandermonde determinant $\det (z_i^{n-j})$ has squared torus integral $n!$, by Fourier orthogonality. Multiplying an irreducible character of highest weight $\lambda$ by this determinant gives an alternating Laurent polynomial. Its leading alternant has coefficient one; character orthogonality for compact groups forces every other alternant coefficient to vanish. This yields the quotient $\det (z_i^{{\lambda }_j+n-j})/\det (z_i^{n-j})$.

For mostly-minus metric signature, the Weyl sigma matrices are ${\sigma }^{\mu }=(I,{\sigma }_1,{\sigma }_2,{\sigma }_3)$ and ${\bar{\sigma }}^{\mu }=(I,-{\sigma }_1,-{\sigma }_2,-{\sigma }_3)$, where ${\sigma }_i$ are the Pauli matrices. The Pauli matrix multiplication law provesA real Lorentz four-vector is equivalently a Hermitian matrix $X=x^{0}I+x^i{\sigma }_i$ with $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$. The transformation $X\mapsto MX{M}^{†}$ for $M\in \mathrm{SL}(2,\mathbb{C})$ preserves this determinant and gives the Lorentz spinor double cover. The indices ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ therefore convert a vector index into one undotted and one dotted Weyl spinor index. They are the invariant tensors in the mixed supercharge anticommutator $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$.

For two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.

For a fixed homogeneous absorber in a narrow-beam geometry, define the X-ray spectral attenuation operator on L1 space over a bounded positive energy interval. A physical input $f(E)\ge 0$ is an integrable function giving incident photon rate density. Let its linear attenuation coefficient $\mu (E)\ge 0$ be a measurable function, with $x\ge 0$. The transmitted uncollided density isHere $w(E)\ge 0$ is a fixed bounded detector response described by a measurable function; background has been subtracted. This is a positive linear operator on L1 space. Since $0\le e^{-\mu (E)x}\le 1$, it cannot increase the integrated absolute incident rate. Multiplication of exponentials proves $A_{x+y}=A_x{A}_y$ for successive layers of the same material. Ignore secondary radiation entering the detector, saturation and changes to the source. Discrete spectral lines use the finite-sum version below instead of an ordinary rate density.

For a chosen filter, stipulate ${\mu }_{\alpha }=0.20\mathrm{m}{\mathrm{m}}^{-1}$, ${\mu }_{\beta }=0.80\mathrm{m}{\mathrm{m}}^{-1}$ and input line rates $(N_{\alpha },N_{\beta })=(4,1)$ in a common arbitrary rate unit. Assume equal calibrated detector weight for both lines. At $x=2.0\mathrm{mm}$,The selected $8.5\mathrm{keV}$ line's fraction rises from $4/5=0.800$ to $0.929974$, but its absolute rate falls by about $33.0\%$. Filtering improves purity at the cost of signal. It multiplies existing components and cannot create a new emitted spectral line.

With just one integrated reading, $S=N_{\alpha }+N_{\beta }$, the two unknown line rates are underdetermined. If the source is unchanged between an empty-path reading and a filtered reading, the ideal inverse problem becomesThe determinant $e^{-1.6}-e^{-0.4}\simeq -0.468424$ is nonzero, so these two ideal readings determine the two stipulated line rates uniquely. Identical coefficients would give zero determinant and destroy that separation. This small linear algebra example explains why absorber contrast supplies information, while also showing its limit: unknown scattered components or unknown detector weights add unknowns that two readings cannot determine.

An invertible matrix need not admit the unpivoted Thomas algorithm. For example,Here $\det A=-1$ and $Ax=d$, but $p_1=0$ makes $m_2=1/0$ undefined. Interchanging the two equations instead gives the upper triangular matrix system $x_1+x_2=3$, $x_2=2$, solved immediately by back substitution. Even a nonzero first diagonal does not guarantee all later pivot elements: $\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\\ 0& 1& 1\end{array}\right)$ has determinant $-1$, yet $p_2=1-1=0$.

For a general tridiagonal matrix, use a solver with partial pivoting when these structural guarantees are unavailable or a dangerously small pivot element occurs. Adjacent row interchanges can introduce a second upper diagonal, so a pivoted solver must track that additional fill-in; it cannot blindly continue the unpivoted recurrence relations. LAPACK's DGTSV explicitly performs partial pivoting, stores this extra upper diagonal, and includes it in back substitution. Its fixed-width storage and bounded work per row retain $O(n)$ time for one right-hand side and $O(n)$ storage.^[ref]