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An $R$-module $M$ is tensor-nilpotent when $M^{\otimes k}=0$ for some positive integer $k$. A nonzero finitely generated module cannot be tensor-nilpotent: localize at a maximal ideal in its support and reduce modulo that maximal ideal, obtaining a nonzero tensor power of a nonzero vector space.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.

The subspace quiver with three sources reduces to the problem of three subspaces in one vector space. The three-subspace decomposition gives eleven indecomposables with vertex dimensions at most one, and one exceptional indecomposable of the three-subspace quiver.

A quantum field theory assigning vector spaces to closed spatial manifolds and linear maps to interpolating spacetime cobordisms, compatibly with composition and disjoint union. Its TQFT partition function on $\Sigma \times S^1$ is the trace of the identity of the state space on $\Sigma$, hence its dimension.

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.

Two norms $p,q$ on a vector space give the same topology if and only if $cp(x)\le q(x)\le Cp(x)$ for some positive constants. Indeed, continuity of the identity at zero gives $p(x)<\delta \Rightarrow q(x)<1$. For $x\ne 0$, apply this to $\delta x/(2p(x))$ to obtain $q(x)<2p(x)/\delta$. Reverse the roles of the norms for the lower bound. Conversely the two inequalities compare every translated open ball. This proof uses absolute homogeneity of a norm, not finite dimension.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

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.

For finite field extensions $K\subseteq L\subseteq M$, the degree of a field extension satisfiesThis tower law is a statement about the dimension of the underlying vector spaces.

Choose a basis $u_1,\dots ,u_r$ of $L$ over $K$ and a basis $v_1,\dots ,v_s$ of $M$ over $L$. The products $u_i{v}_j$ span $M$ over $K$: first expand an element using the $v_j$, then expand each coefficient using the $u_i$.

They are also linearly independent. If ${\sum }_{i,j}{a}_{ij}{u}_i{v}_j=0$ with $a_{ij}\in K$, independence of the $v_j$ gives ${\sum }_i{a}_{ij}{u}_i=0$ for every $j$; independence of the $u_i$ then gives every $a_{ij}=0$. There are $rs$ products, proving the formula.

The transcendence degree is the cardinality of a transcendence basis. It counts independent transcendental parameters. In a finitely generated separably generated field extension, it equals the dimension of the Kähler differentials as a vector space over the extension field.

Choosing a new origin $a$ in a vector space transports its operations along $T(x)=x-a$. Since $T(x\oplus y)=T(x)+T(y)$ and $T(k\odot x)=kT(x)$, every vector space axiom transfers through this bijection. The new additive identity is $a$ and the additive inverse of $x$ is $2a-x$. This illustrates why an affine space has no preferred origin even though choosing one produces a vector space.

On an affine space modeled on a vector space, the map $T_v(x)=x+v$ for a fixed vector $v$. Its derivative is the identity, and $T_v{T}_w=T_{v+w}$. Changes between flat coordinates on a translation surface have this form.

Choosing a positive system of a root system for a complex semisimple Lie algebra divides its root spaces into negative and positive ones. Their respective sums ${\mathfrak{n}}^{-}$ and ${\mathfrak{n}}^{+}$, together with the Cartan subalgebra $\mathfrak{h}$, give this direct sum of vector spaces. Both ${\mathfrak{n}}^{\pm }$ are nilpotent Lie algebras, and $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a Borel subalgebra. The Poincaré-Birkhoff-Witt theorem turns this into an ordered vector-space decomposition of the universal enveloping algebra, useful for building Verma modules.

For a linear operator on a real vector space, the identity $T^3=T$ gives an explicit eigenspace decompositionIndeed, the linear operatorssum to $I$. Their images lie respectively in the eigenspaces for $0$, $1$, and $-1$, since $T{P}_0=0$, $T{P}_{+}=P_{+}$, and $T{P}_{-}=-P_{-}$. On each of these eigenspaces, its corresponding $P$ is the identity and the other two vanish, which proves uniqueness of the decomposition. Thus a finite-dimensional tripotent linear operator admits a diagonalization with diagonal entries in $\{0,1,-1\}$. The condition $T^2\ne T$ is equivalent to the presence of a nonzero $-1$ eigenspace, and $T^2\ne I$ is equivalent to a nonzero kernel.

The trivial representation sends every group element to the identity operator. A one-dimensional constant-coordinate line in a permutation representation is an example: relabelling positions leaves a uniform assignment unchanged. A trivial action on a higher-dimensional vector space has many invariant subspaces and is therefore not an irreducible representation.

Let $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.

In ${\mathbb{F}}_2[u,v]/(u,v{)}^2$, the ideals $(u)$, $(v)$ and $(u+v)$ are pairwise incomparable and nonprime. Their union is the ideal $(u,v)$: the three lines cover its two-dimensional vector space over ${\mathbb{F}}_2$. This shows why a primality hypothesis matters in prime avoidance.

In the operator sense, a finite-dimensional group representation is unipotent when every represented element is a unipotent matrix: for each $g$, $\rho (g)-I$ is a nilpotent endomorphism. This terminology is distinct from the special unipotent characters in the representation theory of finite groups of Lie type.

For example, the generator of $\mathbb{Z}$ can act on a two-dimensional vector space by $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. Then $T^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$, and the first coordinate line gives a nonsplit extension of representations of the trivial representation by itself. Over a field of characteristic zero, a unipotent matrix of finite order is the identity, because its minimal polynomial divides both a power of $X-1$ and the separable polynomial $X^m-1$. Thus every unipotent group representation of a finite group is trivial in characteristic zero; in characteristic $p$, the displayed $T$ instead gives a nontrivial group representation of $C_p$.

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.