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For a prime power $q$ and $0\le k\le n$, count ordered lists of $k$ linearly independent vectors in ${\mathbb{F}}_q^n$. The first has $q^n-1$ choices. After $i$ independent choices, their linear span contains $q^i$ vectors, so the next has $q^n-q^i$ choices. Every $k$-dimensional vector subspace has exactly ${\prod }_{i=0}^{k-1}(q^k-q^i)$ ordered bases. Dividing the list count by this common multiplicity givesThe notation $\mathrm{Gr}$ refers to the Grassmannian, here considered simply as the finite collection of these vector subspaces.^[ref]

For $n=4,k=2,q=2$, there are $15\cdot 14=210$ ordered independent pairs. Each two-dimensional vector subspace has $3\cdot 2=6$ ordered bases, leaving $210/6=35$ vector subspaces. There are only $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ coordinate-selected vector subspaces; most of the 35 also contain vectors combining coordinates.

A unique reduced row echelon form gives another count. Its six possible pivot-column patterns have $4,3,2,2,1,0$ freely chosen entries. Each free entry lies in ${\mathbb{F}}_2$, giving $16,8,4,4,2,1$ vector subspaces, respectively. The diagram fixes the zero entries required to make each pivot the first nonzero entry in its row and the only nonzero entry in its column. Different choices describe different row spaces.

For a nonzero finitely generated algebra over a field, choose a finite-dimensional generating vector subspace $V$ containing $1$. The dimension is $\mathrm{GKdim}A={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_k{V}^n)/\log n$. Changing $V$ only rescales the filtration index by bounded factors. A nonzero finite-dimensional algebra has dimension zero; a polynomial ring in $r$ variables has dimension $r$. The zero algebra is often assigned dimension $-\infty$.

For a nonzero right module $M$ that is a finitely generated module, choose a finite-dimensional generating vector subspace $W$ and define ${\mathrm{GKdim}}_{A}M={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_{k}W{V}^n)/\log n$, where $V$ generates the algebra and contains $1$. For a left module use $V^{n}W$. It is independent of these choices and measures polynomial-order growth of a generating filtration of a module.

For a nonempty radially open convex set $C$ disjoint from a vector subspace $U$, there is a nonzero linear functional $L$ whose kernel of a linear map contains $U$ and avoids $C$. Choose $a\in C$ and the Minkowski functional $p$ of $C-a$. On $U+\mathbb{R}a$, set $g(u+ta)=-t$. Disjointness implies $p(u-a)\ge 1$, which gives $g\le p$ for negative $t$; nonnegativity of $p$ gives it for nonnegative $t$. The algebraic Hahn-Banach theorem extends $g$ to $L\le p$. For $x\in C$, $L(x)+1=L(x-a)<1$, so $L(x)<0$. Also $L(a)=-1$ and $L{|}_U=0$.

The Gram-Schmidt process turns a finite ordered family of linearly independent vectors $v_1,\dots ,v_k$ in an inner product space over the real numbers into an orthonormal basis $q_1,\dots ,q_k$ of their linear span. It keeps every successive linear span, so the first $j$ new vectors describe exactly the same directions as the first $j$ inputs. Its idea is geometric: remove the orthogonal projection onto directions already found, then divide the remaining vector by its norm. This makes a complicated basis easier to use without changing the vector subspace it describes.^[ref]

Start with $u_1=v_1$ and $q_1=v_1/\Vert v_1\Vert$. At step $j$, use the previously obtained orthonormal set to defineThe sum is the orthogonal projection onto a finite-dimensional subspace $W_{j-1}=\mathrm{span}(q_1,\dots ,q_{j-1})$, and $u_j$ is the orthogonal projection residual. Linear independence ensures $u_j\ne 0$: otherwise $v_j$ would already be a linear combination of earlier inputs. If the input family is linearly dependent, use only the retained orthonormal vectors in the sum and omit an input whenever its remainder is the zero vector; do not divide by zero. The retained outputs still form an orthonormal basis of the original linear span, but their number can be smaller than $k$.

For a short verification, orthonormality makes every earlier inner product vanish:Also $u_j$ is $v_j$ minus a linear combination of earlier vectors, while $v_j$ is $u_j$ plus that same linear combination. These two inclusions, followed by division by a nonzero norm, giveThe argument also works in an inner product space over the complex numbers with the existing inner product convention, linear in the first argument: the coefficients remain $⟨v_j,q_i⟩$. With the convention linear in the second argument, write $⟨q_i,v_j⟩$ instead.

For a first picture, take $v_1=(1,1{)}^T$ and $v_2=(1,0{)}^T$ in ${\mathbb{R}}^2$ with the ordinary dot product. ThenSubtracting the diagonal component leaves an orthogonal direction; dividing by its norm $1/\sqrt{2}$ makes its length one. If the second input were $(2,2{)}^T$ instead, its orthogonal projection would be the entire input and $u_2=0$, revealing linear dependence.

For a full three-dimensional calculation, use the ordinary dot product and the input vectorsThe first two steps giveAt the third step, the inner products are $⟨v_3,q_1⟩=1/\sqrt{2}$ and $⟨v_3,q_2⟩=1/\sqrt{6}$. ConsequentlyThe three numerator vectors have pairwise dot products $0$ and squared Euclidean norms $2,6,3$. Thus the displayed outputs are indeed an orthonormal basis, and the three nonzero remainders also confirm linear independence of the inputs. Keeping the removed coefficients produces the QR decomposition below.

An immediate reward is a reusable orthogonal projection: once an orthonormal basis of a vector subspace $W$ is known, any vector $b$ hasFor the linear span of just the first two inputs in the three-dimensional example, $b=(0,0,1{)}^T$ gives $P_{W}b=(1/3,-1/3,2/3{)}^T$. Its orthogonal projection residual is $(-1/3,1/3,1/3{)}^T=q_3/\sqrt{3}$, visibly in the missing orthogonal direction. This is the geometry behind the linear least-squares problem.

The formulas above describe exact arithmetic. With floating-point arithmetic, subtracting an orthogonal projection nearly equal to the input can leave a small remainder; dividing by its small norm can magnify existing roundoff error. The Classical Gram-Schmidt and Modified Gram-Schmidt arrangements agree in exact arithmetic but can give different rounded results. Exact linear dependence and a small computed remainder therefore require different judgments in a numerical implementation.^[ref]

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

A linear system is homogeneous when its forcing is zero. Its solution set is the kernel of a linear map, a vector subspace. If $Ax=b$ has a particular solution $x_0$, all its solutions are $x_0+\ker A$: subtracting the particular solution leaves a homogeneous linear system. The rank-nullity theorem gives the dimension of this freedom.

A hyperplane is a vector subspace of codimension one. A projective hyperplane in ${\mathbb{P}}^n$ is the zero set of one nonzero homogeneous linear form.

The image of a linear map $T:V\to W$ is the vector subspace $\mathrm{im}T=\{T(v):v\in V\}$ of its codomain.

The intersection of any family of vector subspaces of a common vector space is the set of vectors belonging to every member of the family. It is again a vector subspace.

An invariant subspace of a linear map $A:V\to V$ is a vector subspace $W\subseteq V$ satisfying $A(W)\subseteq W$. For a group representation, invariance means $\rho (g)W\subseteq W$ for every represented group element $g$.

Invariance of a vector subspace does not mean that each individual vector stays fixed. The zero-sum plane in the three-coordinate example is invariant even though the cyclic permutation rotates its nonzero vectors. By contrast, the constant-coordinate line is pointwise fixed. Distinguishing these statements prevents a symmetry-preserved mode space from being mistaken for a static mode.

A nonzero Lie algebra representation is irreducible when it has no proper nonzero invariant subspace.

Every finite-dimensional vector subspace $E$ of a Banach space $X$ is the range of a bounded linear projection with the displayed norm bound. In maximal inscribed Euclidean or Hermitian ellipsoid coordinates, write the John contact decomposition as ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I_E$. Its contact functionals ${\varphi }_r(x)=u_r^{\ast }x$ have norm one on $E$. Extend them to $X$ by the Hahn-Banach theorem and set $Px={\sum }_r{c}_r{\tilde{\varphi }}_r(x)u_r$. This restricts to the identity on $E$. The contact synthesis operator has Euclidean norm one, the weights sum to $\dim E$, and the Euclidean unit ball is contained in the norm unit ball; these facts give $\Vert Px{\Vert }_E\le \sqrt{\dim E}\Vert x{\Vert }_X$.

The kernel of a linear map $T:V\to W$ is the vector subspace of inputs sent to zero. Linearity makes it a vector subspace: if $T(v)=T(w)=0$, then $T(av+bw)=0$ for all scalars $a,b$. A vector in the kernel represents a change that the map cannot detect, since $T(x+v)=T(x)$ for every input $x$.

For $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ given by $T(x,y,z)=(x+y,z)$, the zero-output equations are $x+y=0$ and $z=0$. ThusIncreasing the first input while decreasing the second by the same amount leaves the output unchanged. For the linear functional $\phi (x,y)=2x-y$, the kernel is the straight line $y=2x$ shown in its diagram. A linear map is injective exactly when its kernel is the zero vector subspace, because $T(v)=T(w)$ is equivalent to $T(v-w)=0$.

The kernel describes invisible changes, rather than all inputs that produce a specified nonzero output. If $T(x_0)=b$, the corresponding fiber of a function is the affine solution space of a linear equation $x_0+\ker T$. The difference of two inputs in that fiber belongs to the kernel; an input in that fiber need not. In particular, for $b\ne 0$ the fiber cannot be a vector subspace, since it does not contain the zero vector.^[ref]

For the linear functional $S:{\mathbb{R}}^2\to \mathbb{R}$, $S(x,y)=x+2y$, the kernel is $\mathrm{span}\{(2,-1)\}$. Each fiber $S^{-1}(\{c\})$ is the parallel straight line $(c,0)+\mathrm{span}\{(2,-1)\}$. The input vector space is two-dimensional, but the output is one scalar: a whole straight line of inputs collapses to one point of the output real line.