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If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ are bases of finite-dimensional vector spaces $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a basis of $V{\otimes }_{F}W$. There is one independent coordinate for every pair of indices.^[ref]

For a proof outline, bilinearity expands each pure tensor asso these products form a spanning set. For linear independence, take the coordinate functionals $e^k,f^{\ell }$ from the two dual bases. The bilinear map $(v,w)\mapsto e^k(v)f^{\ell }(w)$ induces a linear functional ${\lambda }_{k\ell }$ on the tensor product, satisfying ${\lambda }_{k\ell }(e_i\otimes f_j)={\delta }_{ki}{\delta }_{\ell j}$. Applying it to a proposed zero linear combination selects just $c_{k\ell }$, forcing every coefficient to vanish. Thus dimensions multiply, rather than add. This finite-dimensional count includes a zero factor, whose tensor product has dimension zero.

The space of test functions has the locally convex inductive-limit topology of the spaces of smooth functions supported in a fixed compact set $K$. Each ${\mathcal{D}}_K$ uses seminorms $p_m(\phi )={\max }_{|\alpha |\le m}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }$. Sequential convergence means a common compact support and uniform convergence of every derivative. A linear functional on this space is a distribution exactly when its restriction to each ${\mathcal{D}}_K$ satisfies a finite order of a distribution estimate; the order and constant may depend on $K$.

A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

The total differential is the linear map that combines the first-order responses to simultaneous changes in all coordinates. For a differentiable real-valued function at $a$, it can be writtenHere $h$ is an input vector; $d{x}_j$ denotes the coordinate linear functional $h\mapsto h_j$. The coordinate formula and the Fréchet derivative describe the same linear map. The defining approximation isThe little o notation says the remaining approximation error becomes small relative to the input norm, across every direction of approach.^[ref]

For $f=x^2+xy+2y^2$ at $a=(1,1)$, $d{f}_a(h,k)=3h+5k$. A simultaneous change $(h,k)=(0.02,-0.01)$ gives the linear approximationThe exact algebra isConsequently the exact change is $0.0104$, and the approximation error is $0.0004$. If both input changes are halved, the linear prediction becomes $0.005$ and the error becomes $0.0001$: the first-order response halves, while the quadratic error falls by a factor of four. The right panel under directional derivative displays this comparison as the displacement is scaled.

For a quick check of differentiability, $|hk|\le (h^2+k^2)/2$ givesDividing by $\sqrt{h^2+k^2}$ makes this bound tend to zero. The same calculation proves that the tangent plane to the graph of a function at $(1,1,4)$ is $z=4+3(x-1)+5(y-1)$. It approximates the entire local surface, rather than only two coordinate slices.

The total differential is useful for estimating the combined effect of small input changes without reevaluating a complicated function. For example, if $|h|\le 0.02$ and $|k|\le 0.01$, then $|d{f}_a(h,k)|\le 0.11$. The exact remainder bound from the displayed polynomial is at most $0.0008$, so this example has $|\Delta f|\le 0.1108$ throughout that input rectangle. These are deterministic bounds; a variance calculation additionally needs a joint probability distribution and its covariances. A first-order estimate is also local: large changes require control of the remainder, not merely knowledge of the partial derivatives at one point.

A length-$n$, dimension-$k$ binary linear code has full support when no coordinate vanishes on every codeword, equivalently when the union of the supports of a vector of all codewords covers every coordinate. Every coordinate is then a nonzero linear functional on the message vector space ${\mathbb{F}}_2^k$. Translation by a vector on which it is one pairs the zero and one fibers, so exactly $2^{k-1}$ codewords contribute one in that coordinate. Summing these contributions proves the formula. If there are $r$ nonzero columns in a full-rank generator matrix, the same argument gives $r{2}^{k-1}$ instead.

Over $\mathbb{R}$ or $\mathbb{C}$, every linear functional on all $n\times n$ matrices satisfying $f(AB)=f(BA)$ is a scalar multiple of the trace. The matrix units show that the functional vanishes off the diagonal and has equal values on every diagonal unit. The normalization $f(I)=n$ fixes the scalar to one.

A traceless matrix is a square matrix whose trace is zero. Such matrices form the kernel of the trace linear functional.

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces:

For a nonzero vector $v$ and nonzero linear functional $f$ vanishing on $v$, this invertible map fixes the hyperplane $\ker f$ pointwise and has determinant one. Its inverse is $I-vf$, since $(vf{)}^2=0$. A symplectic transvection is obtained by taking $f=cB(-,v)$ for a nondegenerate alternating bilinear form.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.

The weak limit of a sequence in a normed vector space is an element at which every continuous linear functional has the limit of its values along the sequence. Thus the definition is precisely weak convergence to that element.

A map from an open subset of $\mathbb{C}$ into a complex Banach space is weakly holomorphic when every composition with a bounded complex-linear functional is a scalar holomorphic function. Norm holomorphy requires its difference quotients to converge in the Banach norm. Weak holomorphy is initially a condition on scalar observations; the weak holomorphy implies norm holomorphy argument recovers an actual vector derivative. A closed target subspace containing values on a sequence with an interior accumulation point contains every value on the connected domain: annihilate the subspace, apply the scalar identity theorem, and use Hahn-Banach separation.

Every bounded sequence in a Hilbert space has a weakly convergent subsequence. The closed span of a sequence is separable; diagonal convergence of coordinates along an orthonormal basis gives a bounded limiting linear functional, represented by a vector of the Hilbert space. Applied to a bounded family of mollifications, this proves that their strong Lebesgue limit inherits the weak derivative bound.