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A boundary wavelet replaces a translated interior wavelet whose support of a function meets a domain endpoint. Compatible finite boundary refinement matrices preserve orthogonality and localization across scales. If the coarse boundary spaces reproduce polynomials of degree less than $q$, their detail-space orthogonal complements have $q$ vanishing moments.

For a bounded linear operator $T$ between Hilbert spaces, its range is closed exactly when there is $b>0$ with $\Vert Tv\Vert \ge b\Vert v\Vert$ for all $v\in (\ker T{)}^{\perp }$. The restriction to this orthogonal complement is a bounded bijection onto the range, so the bounded inverse theorem proves necessity. Conversely, the bound makes preimages of a Cauchy sequence of image points a Cauchy sequence, proving closedness by completeness.

A vector subspace closed in the norm topology of a Hilbert space is complete in the inherited norm and inner product. Its orthogonal complement gives a decomposition of the ambient Hilbert space into an orthogonal direct sum. This closedness permits orthogonal projection and ensures limits used in range arguments remain in the subspace. Every finite-dimensional vector subspace is closed.

The cokernel records output information that a linear map cannot produce. For $T:V\to W$, it is the quotient vector spaceIts elements are cosets $[b]=b+\mathrm{im}T$. Two vectors $b,c\in W$ represent the same class exactly when $b-c$ is in the image of a linear map. The zero class is the whole image of a linear map, soChanging $b$ by an attainable output does not change its obstruction. This is useful because one can test compatibility before trying to find an input. The quotient vector space construction is described in Axler's §3E. ^[ref]

For the linear map $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ given by $T(x,y)=(x+2y,0)$, the image of a linear map is the horizontal line. All outputs $(b_1,b_2)$ with the same second coordinate have the same cokernel class, because their difference is horizontal. Thus this cokernel is isomorphic to $\mathbb{R}$ via $[(b_1,b_2)]\mapsto b_2$. In particular, $(3,0)$ has zero class and is attainable, while $(3,1)$ has nonzero class and is unattainable.

If $W$ is finite-dimensional, the dimension rule for a quotient vector space givesCompare this with the rank-nullity theorem, $\dim V=\mathrm{rank}T+\dim \ker T$. The kernel counts invisible inputs; the cokernel counts missing output directions. They need not have the same dimension: $B(x,y,z)=(x+2y,0)$ has a two-dimensional kernel and a one-dimensional cokernel. A linear map is surjective exactly when its cokernel is zero.

The cokernel is a quotient, rather than a chosen perpendicular vector subspace. With a finite-dimensional inner product space as output, one may represent each class by its unique component in the orthogonal complement of the image of a linear map. That representation depends on the chosen inner product. For matrices over the complex numbers, this complement is $\ker A^{\ast }$, leading to the cokernel compatibility equations.

The same quotient definition applies to a group homomorphism of abelian groups. For $T:\mathbb{Z}\to \mathbb{Z}$ given by $T(n)=2n$, the cokernel is $\mathbb{Z}/2\mathbb{Z}$. The equation $2n=b$ is solvable exactly when $b$ is an even integer; the two cosets record integer parity. A quotient captures the obstruction here just as it captures a missing output coordinate for a linear map.

For a matrix $A$ over the complex numbers with the standard Hermitian inner product, the linear system $Ax=b$ is solvable exactly whenThe star is the conjugate transpose; for matrices over the real numbers, use the transpose. Necessity follows from $y^{\ast }Ax=(A^{\ast }y{)}^{\ast }x=0$. For sufficiency, $(\mathrm{im}A{)}^{\perp }=\ker A^{\ast }$, and taking the orthogonal complement again recovers $\mathrm{im}A$ in finite dimension. This is the cokernel obstruction expressed as scalar equations. ^[ref]

For $A=\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)$, $\ker A^T=\mathrm{span}\{(0,1)\}$, so the single compatibility equation is $b_2=0$. If it holds, the kernel supplies the remaining input freedom; if it fails, changing the input cannot remove the obstruction.

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

A finite-dimensional complex representation of a compact group is a direct sum of irreducible representations. Average a positive Hermitian inner product over normalized Haar measure. The orthogonal complement of an invariant subspace is then invariant, so induction on dimension gives the decomposition. The same argument works for finite-dimensional real representations with a positive real inner product.

For finite-dimensional rational representations of complex $G{L}_m$ or $S{L}_m$, average a Hermitian inner product over the compact group $U(m)$ or $SU(m)$ using normalized Haar measure. Orthogonal complements become invariant under the compact group and its complexified Lie algebra. The latter generates the complex group, giving invariant complements and hence complete reducibility. This conclusion does not hold for arbitrary affine algebraic groups, such as the additive group.

The cross-product matrix $A\mathbf{x}=\mathbf{n}\times \mathbf{x}$ obeys $A\mathbf{n}=0$ and $A^2=\mathbf{n}{\mathbf{n}}^T-|\mathbf{n}{|}^{2}I$, by the vector triple product. On the orthogonal complement of a nonzero $\mathbf{n}$, this is $-|\mathbf{n}{|}^{2}I$; hence the two remaining complex eigenvalues are $\pm i|\mathbf{n}|$. In an orthonormal basis adapted to $\mathbf{n}$, $A$ is a planar quarter-turn multiplied by $|\mathbf{n}|$, with zero action along its axis.

A real perturbation with zero diagonal and every off-diagonal element $t$ has one eigenvalue $2t$ on the symmetric vector and a twofold eigenvalue $-t$ on its orthogonal complement. It is $t(J_3-I)$, where $J_3$ is the all-ones matrix. Diagonal entries alone do not give first-order energies in a degenerate subspace.

Suppose asset returns have the exact representation $r=a+Bf$, allowing unrestricted long and short portfolios. A position $v$ with $1^{T}v=0$ and $B^{T}v=0$ costs zero and has constant financial payoff $a^{T}v$. Absence of arbitrage therefore forces $a^{T}v=0$, since either sign of $v$ is allowed. The orthogonal complement identity gives $a\in \mathrm{span}(1,\mathrm{col}B)$. Taking expectations proves the displayed pricing relation, with $\lambda$ equal to the coefficients of $B$ in $a$ plus $\mathbb{E}f$. When $[1B]$ has full column rank, ${\lambda }_0$ is the return of a unit-cost zero-exposure portfolio and ${\lambda }_j$ is the expected return of a zero-cost portfolio with exposure one to factor $j$ and zero to the others. Without that rank condition the coefficients need not be unique, and a traded zero-exposure portfolio need not exist.

For the normalized velocity-reset collision operator on the weighted Hilbert space, $L=\Pi -I$ vanishes on the equilibrium direction and equals $-I$ on its orthogonal complement. The spectral gap is exactly one. Conservation of the projected component and the displayed energy identity give exact exponential decay of the remaining component; the full solution need not decay to zero.

Unrestricted minimization of a Rayleigh quotient seeks the ground state. To bound the $k$th excited energy from above, restrict normalized trial wavefunctions to the orthogonal complement of the exact lower-energy eigenspaces. In an exact orthonormal eigenbasis, all remaining expansion coefficients multiply energies at least $E_k$, so their weighted average is at least $E_k$. A known symmetry sector can supply some of these exclusions automatically. Orthogonality only to inaccurate approximate lower states does not automatically retain the same rigorous excited-state upper bound.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

For a finite-dimensional discounted gain random vector $X$, absence of arbitrage means $h\cdot X\ge 0$ almost surely only when $h\cdot X=0$ almost surely. Let $N=\{h:h\cdot X=0\text{ almost surely}\}$ and minimize $F(h)=\mathbb{E}{e}^{-|X{|}^2-h\cdot X}$ on the orthogonal complement $N^{\perp }$. The Gaussian damping makes $F$ finite and differentiable without physical integrability assumptions on $X$. Every nonzero direction in $N^{\perp }$ has negative gains on an event of positive probability; compactness of directions proves coercivity. At a minimizer, differentiation gives $\mathbb{E}[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]=0$. Normalizing the positive weight produces an equivalent martingale measure with integrable gains. This is a constructive one-period proof of the fundamental theorem of asset pricing even on an infinite probability space.

For a nonzero vector $x\in {\mathbb{R}}^n$, put $r=|x|$ and $\hat{x}=x/r$. The gradient of the Euclidean norm is $\nabla r=\hat{x}$. Differentiate its components to obtain the Hessian matrixThe factor $I-\hat{x}{\hat{x}}^T$ is the orthogonal projection matrix onto the orthogonal complement of the radial direction. Thus the Hessian matrix has eigenvalue zero radially and $1/r$ on each tangential direction, explaining the convexity of the Euclidean norm. Taking the trace gives $\Delta r=(n-1)/r$ away from the origin. The Euclidean norm is not differentiable at the origin: its restriction to any line is a nonzero multiple of the absolute value.

For a linear map $T:V\to W$ between finite-dimensional inner product spaces,The first equality follows directly from the defining identity for the adjoint operator; the second follows by taking orthogonal complements. In an infinite-dimensional Hilbert space, the second equality generally requires closure of the image.

For a uniform ball of radius $R>0$ and positive mass $M$, let $G$ be its centre of mass. Spherical symmetry gives the second-moment matrix $\int u{u}^{T}dm=(M{R}^2/5)I$, so the central inertia tensor is $(2M{R}^2/5)I$. The tensor parallel-axis theorem gives the displayed shifted tensor. For $X\ne 0$ its eigenvector along $X$ has eigenvalue $2M{R}^2/5$, while the two-dimensional orthogonal complement has repeated eigenvalue $M(2R^2/5+|X{|}^2)$. Thus it is an isotropic tensor exactly at $P=G$. Displacement along a chosen axis does not change that axis's moment; the transverse moments increase by $M|X{|}^2$.

Consider lines $r=a+sm$ and $r=b+tn$ with nonzero direction vectors. If they intersect, then $a-b=tn-sm$, so the necessary coplanarity condition is $[a-b,m,n]=0$. For nonparallel directions it is also sufficient, since the orthogonal complement of $m\times n$ is the linear span of $m,n$. For parallel directions the condition alone is insufficient: distinct parallel lines also satisfy it.

Assume intersection and $D=[a,m,n]\ne 0$. Take the scalar triple products $[a,b,n]$ and $[a,b,m]$ of $b=a+sm-tn$. The scalar triple product is an alternating multilinear map, giving respectively $sD$ and $tD$, proving the title formulas. The nonzero denominator fixes a convenient determinant chart; intersections with $D=0$ can still exist and require a different auxiliary vector or another method.

For a complex finite-dimensional group representation, the displayed matrices give another group representation: taking inverses and taking the conjugate transpose each reverse multiplication order, so their composition preserves it. This operation is involutive. With the standard Hermitian inner product, an invariant vector subspace $W$ for $D^{♯}$ has orthogonal complement invariant for $D$, because $D(g{)}^{†}W=D^{♯}(g^{-1})W\subseteq W$. Thus $⟨w,D(g)v⟩=⟨D(g{)}^{†}w,v⟩=0$ for $v\in W^{\perp },w\in W$. It follows that $D^{♯}$ is an irreducible representation exactly when $D$ is. No finiteness assumption on the group or unitarity assumption on the group representation is needed.