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If $T=S^{\ast }$ for a bounded operator on a separable Banach space, and $C$ is a nonempty weak-star closed convex subset of the dual unit ball with $T(C)\subseteq C$, then $T$ has a fixed point in $C$. The Cesàro averages of the orbit remain in $C$, and $T$ minus the identity sends those averages to vectors of norm at most $2/n$. Weak-star compactness supplies a cluster point, and the adjoint action is weak-star continuous. The corresponding assertion with only weak closedness is false for the right shift on the probability simplex in ${\ell }^1$, the dual of $c_0$.

The wedge product combines a degree-$p$ differential form and a degree-$q$ differential form into one of degree $p+q$. It is associative and graded-commutative:In particular, $dx\wedge dy=-dy\wedge dx$ and $dx\wedge dx=0$. For planar vectors $v,w$,This determinant is the signed area of their parallelogram. Exchanging the edges reverses the sign; parallel edges give zero. Thus antisymmetry records the geometry of oriented area rather than adding an arbitrary algebraic convention. The wedge product of differential forms requires no Riemannian metric to define it. It obeys the graded Leibniz rule under the exterior derivative.

Linearizing weakly coupled locked phases gives $0=\xi -\Omega 1-kL\theta$, where $L$ is the directed graph Laplacian. Multiplying by a normalized left null vector ${\pi }^{\mathsf{T}}$ eliminates coupling and gives the displayed weighted frequency. In a balanced network, including symmetric reciprocal coupling, $\pi =1/N$ and the frequency is the arithmetic mean. A zero diagonal in the coupling matrix alone does not imply balance. For two directed weights $p,q>0$, the weights are $(q,p)/(p+q)$.

A symmetric positive-definite matrix $M$ defines an inner product on real coordinate vectors byFor diagonal $M$, positive diagonal entries weight the squared coordinate errors. This is the finite-coordinate version of a weighted inner product; the positive-definite matrix example appears in Francis Narcowich, Inner Products and Norms, standard inner products, example 8.

For $M=\mathrm{diag}(4,1)$, the squared norm of $(a,b{)}^T$ is $4a^2+b^2$. A first-coordinate error of one costs as much as a second-coordinate error of two. The unit sphere is the ellipse $4a^2+b^2=1$, and orthogonality means $4u_1{v}_1+u_2{v}_2=0$. Thus “closest” and “perpendicular” depend on the chosen inner product. The orthogonal projection residual example shows how this changes a best approximation.

For a finite undirected graph without self-loops, give each edge a symmetric nonnegative weight $w_{ij}=w_{ji}$. In the weighted graph, set $(A_w{)}_{ij}=w_{ij}$ for an edge and zero otherwise, and $(D_w{)}_{ii}={\sum }_j{w}_{ij}$ with all other entries zero. The Weighted graph Laplacian isEach edge appears once in the quadratic form. For the path graph $A-B-C$ with weights $w_{AB}=2$ and $w_{BC}=3$,The Weighted graph Laplacian is still a positive semidefinite matrix. Its kernel consists of vectors constant on each connected component of a graph after zero-weight edges are removed: an edge of weight zero imposes no constraint. In an electrical network, taking the weights to be electrical conductances makes $x^T{L}_{w}x$ the total power dissipated by Joule heating when $x$ gives the voltages at the vertices. This physical interpretation explains why stronger connections penalize disagreement more heavily.^[ref]

With relaxation parameter $\omega$, weighted Jacobi updates byIts iteration matrix is $H_{\omega }=I-\omega D^{-1}A$.

Equivalently, the weighted Jacobi method takes a weighted average of the old vector and its full Jacobi method update: $x^{(k+1)}=(1-\omega )x^{(k)}+\omega x_{\mathrm{Jacobi}}^{(k+1)}$. If $Bv=\lambda v$, its iteration matrix multiplies that eigenvector by $1-\omega +\omega \lambda$. For the diagonal-4 Jacobi iteration for a three-variable tridiagonal system, the factors are $1-\omega$ and $1-\omega \pm \omega \sqrt{2}/4$; a Poisson equation smoothing parameter must not be inferred from this different matrix.

Take labels $E=\{a,b,c\}$ and vectors in the real vector space ${\mathbb{R}}^2$:Here is the complete independence check for all eight subsets. The empty set is independent. Each of $\{a\},\{b\},\{c\}$ is independent because its vector is nonzero. Every pair is independent: the determinants of the two-column matrices $[v_a{v}_b]$, $[v_a{v}_c]$, and $[v_b{v}_c]$ are respectively $1$, $1$, and $-1$. The triple is dependent because $v_c-v_a-v_b=0$. ThereforeDeleting elements from any listed subset stays in this list, verifying heredity. Every exchange comparison has sizes $(0,1)$, $(0,2)$, or $(1,2)$. From the empty set, any element of the larger matroid independent set works: this covers three singleton targets and three pair targets. From any singleton to any pair, every element of the pair outside the singleton works, since every resulting pair is independent. This covers all nine singleton-pair comparisons, including disjoint ones. There are no other comparisons: all fifteen ordered size-increasing pairs satisfy exchange.

The three matroid bases are $\{a,b\}$, $\{a,c\}$, and $\{b,c\}$, all of cardinality two, with weights $13$, $14$, and $11$. The matroid greedy algorithm considers $a$, then $c$, then $b$: it admits $a$, admits $c$ because their determinant is nonzero, and rejects $b$ because the triple is dependent. The resulting matroid basis $\{a,c\}$ has weight $8+6=14$, the largest of the three totals. Every singleton can grow, so none is maximal; the three pairs are both maximal and maximum in cardinality.

A coordinate matrix expressing restriction of weights to the Cartan subalgebra of a Lie subalgebra. With Dynkin labels as column vectors, its rows encode the subalgebra's coroots in terms of the original coroots. It need not be an idempotent projection operator: an equal-rank embedding can give an invertible matrix. Applying it to the full weight system, including weight multiplicities, permits reconstruction of the restricted irreducible representations.

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.

Give the four corners of the square the exclusive or labels: $+1$ when the input signs differ and $-1$ when they agree. In the following order,No affine function $g(x)=a_1{x}_1+a_2{x}_2+b$ has the required strictly positive scores at $p_2,p_3$ and strictly negative scores at $p_1,p_4$: both pairs have midpoint zero, so $g(p_2)+g(p_3)=2b=g(p_1)+g(p_4)$ would have to be both positive and negative. This also rules out perfect classification by a single threshold with a consistent tie rule.

Under the polynomial kernel's quadratic feature map, the two classes become two distinct vectors:The hyperplane ${\varphi }_2=0$ separates them. With $w=(0,-1/\sqrt{2},0{)}^T$ and $b=0$, the score and classifier areAll four signed scores $y_{i}f(p_i)$ equal one. In the input plane the boundary is $x_1{x}_2=0$, the union of the coordinate axes; in the derived coordinates it is a single hyperplane. The feature map identifies opposite corners, which is harmless for these labels because opposite corners belong to the same class. A fixed feature map would not separate labels that disagree at identical mapped vectors.

The zero vector of a vector space is its additive identity: $v+0=0+v=v$ for every vector $v$. It is unique, because if $0^{\prime }$ is another such identity then $0=0+0^{\prime }=0^{\prime }$. Distributivity gives $\alpha 0=\alpha (0+0)=\alpha 0+\alpha 0$, so $\alpha 0=0$ after cancellation. In a coordinate vector space it has every coordinate zero.