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The underlying graph of a finite Dynkin diagram is a forest. Normalize every simple root to length one. Each bonded pair has inner product at most $-1/2$, and every other distinct pair has nonpositive inner product. A cycle with $m$ vertices would make the squared norm of their sum at most $m-m=0$, contradicting their linear independence in an inner product space. Multiple bonds denote root-length information, not graph-theoretic parallel-edge cycles.

On a complex inner product space $V$, with the inner product conjugate-linear in its first argument, an everywhere-defined linear map $T:V\to V$ admits an adjoint on $V$ if there is a linear map $T^{\ast }:V\to V$ satisfyingThe adjoint, when it exists, is unique: the difference of two candidate values at $y$ is orthogonal to every vector and hence to itself. In a finite-dimensional vector space, it always exists, and its matrix in an orthonormal basis is the conjugate transpose. In a space that is not complete, existence must instead be checked.

The identity $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y\in V$ supplies such an adjoint directly, namely $T^{\ast }=T$. This algebraic statement should be distinguished from self-adjointness of a linear operator on a completed Hilbert space, which additionally compares the domains of an operator and its Hilbert-space adjoint. The Gaussian polynomial operator is a useful example where the identity holds on the entire polynomial inner product space and can be used on every finite-dimensional invariant polynomial subspace.

A matrix whose columns are the coordinate vectors of an ordered basis in a fixed reference basis. It is an invertible matrix, and multiplication by it converts coordinates in the column basis to the reference coordinates: $v=B[v{]}_B$. For a real inner product space with orthonormal reference basis, its Gram matrix is $B^{T}B$, and its reciprocal basis has basis matrix $B^{-T}$.

For three vectors $a,b,c$ in an inner product space, expansion of each squared norm provesTherefore the second-order backward differentiation formula $3U^{n+2}-4U^{n+1}+U^n=2kL{U}^{n+2}$ with a dissipative operator $L$ decreases the discrete energy ${\mathcal{E}}_n=\Vert U^{n+1}{\Vert }^2+\Vert 2U^{n+1}-U^n{\Vert }^2$. This quadratic form is equivalent to the product norm of the two time levels with constants independent of $k$ and $L$. It proves uniform stability even when amplification roots of a polynomial coalesce inside the unit disk, where an eigenvector-separation proof may lose its bound.

On the finite formal span of symbols $e_x$, define $B(e_x,e_y)=k(x,y)$ and extend bilinearly. For a positive-semidefinite kernel, the null vectors of this positive semidefinite bilinear form form its radical of a bilinear form. Quotienting by that linear subspace gives an inner product space and feature map $x\mapsto [e_x]$ with $⟨[e_x],[e_y]⟩=k(x,y)$. This construction works for degenerate and zero kernels.

In an inner product space, the Cauchy-Schwarz inequality isFor $v\ne 0$, subtract the orthogonal projection of $u$ onto $v$ and use nonnegativity of its squared norm; equality means that remainder vanishes. Thus equality holds exactly for linearly dependent vectors, including a zero vector. Applied to square-integrable random variables, it gives $|\mathbb{E}[XY]|\le \sqrt{\mathbb{E}[X^2]\mathbb{E}[Y^2]}$; centring gives the bound $|\mathrm{Cov}(X,Y)|\le {\sigma }_X{\sigma }_Y$.

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.

The covolume of a full Euclidean lattice is the Lebesgue measure of a fundamental parallelepiped in its ambient inner product space. In orthonormal coordinates it is $|\det B|$ for a lattice basis matrix $B$. Scaling the lattice by $t>0$ multiplies the covolume by $t^n$; the dual lattice has reciprocal covolume.

A finite root system in a real inner product space is crystallographic when every Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer. The roots of a complex semisimple Lie algebra satisfy this because each number is a weight of its sl2 subalgebra associated with a root. Reducedness is a separate axiom; the root-space reducedness lemma establishes it for these Lie-algebra roots.

For a unit vector $u$ in a real inner product space, stretching by a factor $\alpha$ along $u$ and leaving its perpendicular subspace fixed givesDecompose $x=(u^{T}x)u+(x-(u^{T}x)u)$. The first term is the orthogonal projection onto the line; the second is perpendicular to it. The displayed map multiplies the first by $\alpha$ and fixes the second. In the plane, if $u=(\cos \theta ,\sin \theta {)}^T$, then $T=R(\theta )\mathrm{diag}(\alpha ,1)R(-\theta )$.

A unital subalgebra $A\subseteq \mathrm{End}(V)$ closed under the adjoint operator, on a finite-dimensional real or complex inner product space, equals its double commutant of an operator algebra. Here is a proof that does not require a classification of algebras. For $N=\dim V$, choose a basis $(v_1,\dots ,v_N)$ and let $W=\{(a{v}_1,\dots ,a{v}_N):a\in A\}\subseteq V^N$. The diagonal action of $A$ preserves $W$ and $W^{\perp }$, so the orthogonal projection onto $W$ commutes with that action. Each matrix entry of this projection belongs to $A^{\prime }$. Hence the diagonal action of $b\in A^{\prime \prime }$ commutes with the projection and preserves $W$. Since $1\in A$, the tuple $(v_1,\dots ,v_N)$ belongs to $W$, and therefore $(b{v}_1,\dots ,b{v}_N)=(a{v}_1,\dots ,a{v}_N)$ for one $a\in A$. This implies $b=a$. Without the unital hypothesis the conclusion is $A^{\prime \prime }=A+\mathbb{F}1$.

For $\lambda \ge 0$ and a linear map $A$ between finite-dimensional real inner product spaces,Thus the penalty is the support function of the compact convex set $A^{\ast }\{p:{\max }_i\Vert p_i{\Vert }_{q^{\prime }}\le \lambda \}$. The Moreau decomposition makes its proximal operator the identity minus the Euclidean projection onto a convex set.

For an orthogonal projection $P$ onto $W$ in an inner product space, apply the Cauchy-Schwarz inequality to $Pu,Pv$ and then use $\Vert Pu\Vert \le \Vert u\Vert$, $\Vert Pv\Vert \le \Vert v\Vert$. If either original vector is zero, equality holds. Otherwise equality holds exactly when both vectors belong to $W$ and are linearly dependent. Necessity follows because equality in the product of the two norm bounds forces equality in each, while equality in Cauchy-Schwarz inequality forces dependence. This retains the zero cases that a division by the original norms would lose.

A Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]

The magnitude of a vector sum is generally different from the sum of the magnitudes. Here $\Vert \mathbf{u}+\mathbf{w}{\Vert }_2=5$, whereas $\Vert \mathbf{u}{\Vert }_2+\Vert \mathbf{w}{\Vert }_2=\sqrt{5}+\sqrt{10}>5$. The triangle inequality explains why the direct displacement cannot exceed the length of this two-segment route. A return displacement $(-3,-4)$ makes the net vector sum zero, although the lengths of the three travelled segments sum to a positive number.

The dot product exposes direction quantitatively: for nonzero Euclidean vectors, $⟨\mathbf{u},\mathbf{w}⟩=\Vert \mathbf{u}{\Vert }_2\Vert \mathbf{w}{\Vert }_2\cos \theta$. For the displayed pair, the dot product is $2\cdot 1+1\cdot 3=5$, so $\cos \theta =1/\sqrt{2}$ and the angle is $45^{\circ }$. For orthogonal vectors, that dot product is zero. Dividing by magnitudes separates direction from size, which is the purpose of vector normalization.

On a finite-dimensional real inner product space, a skew-adjoint operator has even rank. In an orthonormal basis its matrix is skew-symmetric, so an invertible such matrix must have even dimension by $\det A=\det (-A)=(-1{)}^d\det A$. Moreover $\ker A=(\mathrm{im}A{)}^{\perp }$, so the image and kernel intersect trivially. The restriction of $A$ to its image is an invertible skew-adjoint map, and its dimension, the rank, is therefore even.

For a real inner product space $V$, the displayed map identifies ${\Lambda }^{2}V$ with its Special orthogonal Lie algebra. It sends a wedge of distinct orthonormal basis vectors to the corresponding elementary skew-symmetric matrix, so it is an isomorphism. For an orthogonal transformation $g$, invariance of the inner product gives $\Phi (gu\wedge gv)=g\Phi (u\wedge v)g^{-1}$. Thus the exterior square of the defining representation is the Adjoint representation of the orthogonal group. Complexification identifies their weight spaces, allowing the root system to be read from the exterior-square weights and the zero-weight multiplicity.

A finite family of pairwise commuting normal operators on a finite-dimensional complex inner product space, commuting also with each other's adjoints, generates a commutative unital C-star algebra. Its finite maximal ideal space identifies the algebra with functions on finitely many points. The characteristic functions of those points pull back under the Gelfand transform to mutually orthogonal self-adjoint linear projections summing to the identity. Their ranges are the simultaneous eigenspaces, giving simultaneous diagonalization without choosing separate incompatible eigenbases.

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.

A mapping $T$ on an inner product space is firmly nonexpansive whenEvery firmly nonexpansive mapping is nonexpansive, and the resolvent of a maximal monotone operator is firmly nonexpansive.

For vectors $v_1,\dots ,v_n$ in a real or complex inner product space, averaging over all $2^n$ sign choices givesExpand the squared norm. Each diagonal term is $\Vert v_j{\Vert }^2$ in every summand. Each mixed term contains a factor ${ϵ}_j{ϵ}_k$ with $j\ne k$; reversing ${ϵ}_j$ pairs it with an equal term of opposite sign. Thus all mixed terms cancel. Equivalently, independent symmetric random signs have $\mathbb{E}[{ϵ}_j{ϵ}_k]=0$ for distinct indices.

For $n=2$ this is the parallelogram law $\Vert u+v{\Vert }^2+\Vert u-v{\Vert }^2=2\Vert u{\Vert }^2+2\Vert v{\Vert }^2$. It expresses a special feature of inner product geometry: the average squared length of a random signed sum is just the sum of the individual squared lengths.