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The square-class group identifies two nonzero elements of a field when their quotient is a square. Multiplication makes it an abelian group of exponent two, hence a vector space over ${\mathbb{F}}_2$. It records quadratic extensions and the images of Kummer theory for $n=2$.

For a linear operator on a finite-dimensional vector space over a field $K$, all polynomial kernels stabilize at their first power exactly when its minimal polynomial $m$ has no repeated irreducible polynomial factor in $K[X]$. This formulation does not require those factors to be linear.

To prove it, the Bézout's identity for polynomials gives $\ker q(T)=\ker \gcd (q,m)(T)$: a greatest common divisor is a combination of $q$ and $m$, while $q$ is its multiple. If $m$ is square-free, $\gcd (p^n,m)=\gcd (p,m)$, proving the displayed equality. Conversely, if an irreducible factor $f$ occurs to exponent $e\ge 2$, its primary decomposition theorem summand has minimal polynomial $f^e$. On this summand $f(T)$ is nonzero, while $f(T{)}^e=0$. Thus the kernels of those two powers differ. An eigenbasis follows from this criterion precisely when the distinct factors of $m$ are also linear over $K$; a real rotation through a right angle has square-free $m=X^2+1$ but no real eigenvalues.

A matrix with the same number of rows and columns represents an endomorphism of a finite-dimensional vector space after choosing a basis. Square matrices support trace, determinant, powers and the matrix exponential.

An element of an associative algebra is square-zero if its square vanishes. Every such element is a nilpotent element. These elements are preserved by an algebra isomorphism, so their structure distinguishes algebras. Their set need not be a vector space: the matrix units $E_{12}$ and $E_{21}$ are square-zero, while their sum has square equal to the identity in $M_2(K)$. In a square-zero ideal, by contrast, every pairwise product vanishes and all its elements are square-zero. In $K[t]/(t^3)$ the square-zero elements are exactly the multiples of $t^2$, over every field: $(yt+z{t}^2{)}^2=y^2{t}^2$ forces $y=0$.

For a linear endomorphism $h$ of an $n$-dimensional vector space, the images of a linear map formed by its powers decrease and stabilize by exponent $n$. Indeed, $h^{m+k}v=h^m(h^{k}v)$ proves inclusion. If equality occurs at one step, applying $h$ to both sides proves equality at the next step, and induction gives equality forever after. Before stabilization, every inclusion is strict and lowers dimension by at least one. Starting at dimension $n$ allows at most $n$ such decreases.

By the rank-nullity theorem, the increasing null spaces of the same powers stabilize at the same time. This gives a short proof of the finite-dimensional Fitting decomposition.

The standard basis of the vector space ${\mathbb{R}}^n$ consists of the vectors $e_i$ whose $i$th coordinate is one and whose other coordinates are zero. Every vector $x$ has the unique expression $x={\sum }_i{x}_i{e}_i$ in this basis.

The standard basis vector $e_i$ in ${\mathbb{F}}^n$ has coordinate one in position $i$ and zero elsewhere. Every vector $v=(v_1,\dots ,v_n)$ has the unique expansion $v={\sum }_{i=1}^n{v}_i{e}_i$.

For example, $e_2=(0,1,0{)}^{\top }$ in three dimensions. The outer product $e_i{e}_j^{\top }$ is the matrix unit with a single one in row $i$, column $j$, providing a standard basis for matrices as well. The word standard refers to the chosen coordinate representation; an abstract vector space does not come with a distinguished coordinate basis automatically.

A subset $S$ of a real vector space is a star-shaped set if some point $p\in S$ sees every point of $S$ along a segment lying in $S$: $(1-t)p+tx\in S$ for all $x\in S$ and $0\le t\le 1$. Every convex set is star-shaped about each of its points; a star-shaped set need only have one such point.

For example, the union of the horizontal and vertical line segments through the origin is star-shaped about the origin but is not convex. In a topological vector space, the map $H(x,t)=(1-t)x+tp$ contracts a star-shaped set to $p$. This makes it contractible, a useful bridge from elementary geometry to homotopy theory.

If a vector space is spanned by $n$ vectors, every linearly independent family in it has at most $n$ members. Consequently, any two finite bases have the same number of vectors.

For vector subspaces $U\subseteq W\subseteq V$ of a finite-dimensional vector space, such a $W^{\prime }$ always exists. Extend a basis of $U$ to one of $W$, then extend again to one of $V$. Let $W^{\prime }$ be spanned by the original basis of $U$ together with the vectors added in the second extension. Uniqueness of coordinates gives the displayed intersection and sum. Equivalently, choose a direct sum complement $C$ with $V=W\oplus C$ and put $W^{\prime }=U\oplus C$. Necessarily $\dim W^{\prime }=\dim V-\dim W+\dim U$. In $M_2(\mathbb{R})$, the skew-symmetric subspace $U$ inside the traceless subspace $W$ admits $W^{\prime }=\mathrm{span}\{I,\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\}$.

A super vector space is a vector space graded by the two-element group. Degree-preserving maps and the usual graded tensor product form a monoidal category. Over a field of characteristic different from two, this category has two different symmetries: ordinary interchange of tensor factors and interchange multiplied by $(-1{)}^{pq}$ on homogeneous factors of degrees $p,q$. The latter is the Koszul sign rule.

Over a field of characteristic different from two, every square matrix decomposes uniquely asHere $S$ is a symmetric matrix and $K$ is a skew-symmetric matrix. Transposing proves those properties. For uniqueness, a matrix in both classes satisfies $M=M^T=-M$, hence $2M=0$ and $M=0$. This is a direct sum decomposition of the vector space of square matrices. In characteristic two, division by two is unavailable and the two classes coincide, so the same decomposition statement fails.

Let $S:W\times W\to W$ be a symmetric bilinear map on a finite-dimensional real vector space. If $S(v,v)$ is parallel to $v$ for every $v$, then it has the displayed form for one covector $V$. Write $S(e_i,e_i)={\alpha }_i{e}_i$ in a basis. Applying the condition to $e_i+e_j$ and $e_i-e_j$ excludes every component outside their span and gives $S(e_i,e_j)=({\alpha }_j{e}_i+{\alpha }_i{e}_j)/2$. Set $V(e_i)={\alpha }_i/2$ and extend by bilinearity. Dimension one follows by the same definition. The trace identity $S^a{}_{ac}=(n+1)V_c$ proves uniqueness without depending on the chosen basis.

For a finite-dimensional symplectic vector space, the symplectic adjoint of a linear map $T$ is the unique linear map $T^{\ast }$ satisfying the displayed identity. Existence and uniqueness follow because the nondegenerate bilinear form identifies the vector space with its dual space. If $\omega$ has matrix $J$ in a basis, then $T^{\ast }=J^{-1}{T}^{\mathsf{T}}J$. Thus $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$ and $(T^{\ast }{)}^{\ast }=T$, and membership in the symplectic Lie algebra is equivalent to $T^{\ast }=-T$. This operation is linear over the ground field; it does not involve complex conjugation. It preserves diagonalizability and the property of being a nilpotent linear map, since it is conjugate to matrix transpose.

For a finite-dimensional vector space over any field with a nondegenerate alternating bilinear form $B$, the symplectic group over a field consists of invertible linear maps preserving $B$. In a symplectic basis its matrices satisfy $g^{\mathsf{T}}Jg=J$. Matching two symplectic bases gives a unique such linear map, and a symplectic basis exists by successively splitting off paired vectors. This includes arbitrary-characteristic and infinite-field versions, beyond the real Lie group and the symplectic group over a finite field.

A symplectic vector space is a vector space equipped with a nondegenerate bilinear form $\omega$ that is an alternating bilinear form: $\omega (v,v)=0$ for every vector $v$. This implies $\omega (v,w)=-\omega (w,v)$. When the characteristic is two, that latter equation alone does not imply the alternating condition. A symplectic vector space of finite dimension has even dimension: splitting off a hyperbolic pair leaves a smaller nondegenerate bilinear form, and repeating gives a decomposition into two-dimensional planes.

For an embedded regular surface $M\subseteq {\mathbb{R}}^3$ and $p\in M$, the affine tangent plane isIt is a set of points through $p$, whereas the tangent space $T_{p}M$ is a vector space of displacements with zero vector. Some texts also call that two-dimensional tangent space the tangent plane; here the translation by $p$ distinguishes the two uses. For a regular parametric surface $X(u,v)$ with $p=X(u_0,v_0)$,where the partial derivatives are evaluated at $(u_0,v_0)$. Regularity means that these tangent vectors are linearly independent, equivalently $X_u\times X_v\ne 0$. Their cross product is a normal vector. ^[ref]

For an implicit regular surface $F(x,y,z)=0$, require a smooth function $F$ with $\nabla F(p)\ne 0$. ThenDifferentiating $F(\gamma (t))=0$ proves that every surface velocity lies in the displayed kernel of a linear map; the implicit function theorem gives a local graph with two independent coordinates, proving equality. The gradient is the normal vector with respect to the ambient dot product. ^[ref]

For the sphere $F(r)=r\cdot r-R^2$, $\nabla F(p)=2p\ne 0$, and the affine tangent plane has equation $p\cdot r=R^2$. At the unit sphere's north pole $p=(0,0,1)$, compare the plane $z=1$ with the displacement tangent space $v_z=0$. The regular parametric surface $X(u,v)=(u,v,\sqrt{1-u^2-v^2})$, for $u^2+v^2<1$, gives $X_u(0,0)=(1,0,0)$ and $X_v(0,0)=(0,1,0)$, confirming the same answer.

This analytical illustration shows a unit sphere and a vertical cross-section. The smooth curves ${\gamma }_1(t)=(\sin t,0,\cos t)$ and ${\gamma }_2(t)=(0,\sin t,\cos t)$ give the two displayed tangent vectors at $p$. The straight point $p+tv$ for any nonzero $v\in T_p{S}^2$ is outside the sphere when $t\ne 0$, since $\Vert p+tv{\Vert }^2=1+t^2\Vert v{\Vert }^2$. A tangent plane therefore captures first-order motion, not an exact patch of the regular surface.

The same idea makes local approximation useful. For the graph of a function $z=f(x,y)$ with $f$ a smooth function, the affine tangent plane is $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$. ^[ref] With continuous second partial derivatives, the vertical error is $O((x-a{)}^2+(y-b{)}^2)$ in Big O notation. On the unit sphere near its north pole, $\sqrt{1-u^2-v^2}=1-\frac{1}{2}(u^2+v^2)+O((u^2+v^2{)}^2)$: the tangent plane loses quadratic bending while preserving all first-order velocities.

A failed regularity test needs interpretation. Replacing the sphere equation by $G(r)=(r\cdot r-1{)}^2=0$ makes $\nabla G=0$ everywhere on the same perfectly regular sphere, so this defining smooth function cannot recover its tangent plane. Similarly, the usual spherical parametric surface loses matrix rank at the poles, while the graph parametrization above remains regular there. A vanishing gradient or degenerate parametrization invalidates that calculation; it alone does not prove that the underlying geometry has no tangent plane.

At a fixed point $p$ of an $n$-dimensional smooth manifold $M$ without boundary, the tangent space $T_{p}M$ is the real vector space of all tangent vectors at $p$. In manifold charts $x=(x^1,\dots ,x^n)$ and $y=(y^1,\dots ,y^n)$, a tangent vector has componentsThe coordinate basis transforms inversely, so the tangent vector is unchanged. This invertible Jacobian matrix also preserves vector addition and scalar multiplication, giving a vector space of dimension $n$ independently of coordinates. ^[ref]

For example, at $(x,y)=(1,0)$ in ${\mathbb{R}}^2$, use the alternative manifold chart $u=x$, $w=y+x^2$. The Jacobian matrix sends $(v_x,v_y)=(2,3)$ to $(v_u,v_w)=(2,7)$. Indeed, the smooth curve $(x(t),y(t))=(1+2t,3t)$ gives $w(t)=1+7t+4t^2$. Conversely, ${\partial }_u{|}_p={\partial }_x{|}_p-2{\partial }_y{|}_p$ and ${\partial }_w{|}_p={\partial }_y{|}_p$, so $2{\partial }_u{|}_p+7{\partial }_w{|}_p=2{\partial }_x{|}_p+3{\partial }_y{|}_p$. The numbers change; their action on every smooth function does not.

For an embedded submanifold $M\subseteq {\mathbb{R}}^N$, the tangent space can be identified with the vector subspaceA proof outline explains why every coordinate vector is possible: in a manifold chart $x$, lift the short straight path $x(p)+ta$ using $x^{-1}$. Its tangent vector has components $a$. For a regular local parametrization $X:U\subseteq {\mathbb{R}}^n\to M$, the same chain rule gives $T_{p}M=\mathrm{im}D{X}_q$, where $X(q)=p$. This is the tangent space from a local parametrization.

For the sphere $S_R^2=\{p\in {\mathbb{R}}^3:p\cdot p=R^2\}$, differentiating that equation gives $p\cdot v=0$. Conversely, for any nonzero $v$ with $p\cdot v=0$, the smooth curvelies on the sphere and has ${\gamma }_v^{\prime }(0)=v$; use a constant smooth curve for $v=0$. Thus $T_p{S}_R^2=\{v:p\cdot v=0\}$, not merely a subset of that plane. At $p=(0,0,R)$ this vector subspace is $v_z=0$, while the affine tangent plane drawn through $p$ is $z=R$.

This also explains a mechanical constraint without requiring a force calculation. In the ideal mathematical model of a particle confined to a stationary sphere, its allowed velocities at $p$ are exactly $T_p{S}_R^2$: the radial component must vanish, but either independent tangential direction is available. For a general stationary holonomic constraint $F(q)=0$ with independent equations, the regular level set theorem gives $T_{q}M=\ker D{F}_q$. These are instantaneous kinematic possibilities; they do not determine the actual subsequent motion. On a general smooth manifold, tangent spaces at different base points cannot be identified canonically; an affine connection supplies parallel transport when such comparisons are needed.

A rank-$n$ tensor on a finite-dimensional vector space $V$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. Its components transform by one copy of the change-of-basis matrix for each index.

The tensor algebra of a vector space $V$ is the graded algebrawith multiplication given by concatenation of tensors.