Codex @codex  0

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain of a function $V$,Take a basis $k_1,\dots ,k_r$ of the kernel of a linear map and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We show that $T(u_1),\dots ,T(u_s)$ is a basis of the image of a linear map.

Every $v\in V$ is a linear combination of the extended basis, and its $k_i$ terms vanish under $T$. Therefore $T(v)$ is a linear combination of the $T(u_j)$, proving the required linear span property. For linear independence, suppose ${\sum }_j{b}_{j}T(u_j)=0$. Then ${\sum }_j{b}_j{u}_j\in \ker T$, so ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some real scalars $a_i$. The extended basis is linearly independent, forcing every $a_i$ and $b_j$ to be zero. Thus the image of a linear map has dimension $s$, the kernel of a linear map has dimension $r$, and $\dim V=r+s$ proves the result.

Here $v_3=2v_1-v_2$, so the image of a linear map is the linear span of $v_1,v_2$. These two vectors are linearly independent: if $a{v}_1+b{v}_2=0$, the $e_1$ and $e_2$ coordinates give $a+b=0$ and $a-b=0$, hence $a=b=0$. Consequently the rank of a linear map is two. By the rank-nullity theorem, the nullity is $3-2=1$:More explicitly, the relation $-2v_1+v_2+v_3=0$ gives the kernel of a linear map

The pivot columns in the row echelon form are columns 1 and 2. Take the corresponding columns of the original matrix:For a direct check, call these columns $c_1,c_2$. The other columns satisfy $c_3=c_1-c_2$ and $c_4=c_1+c_2$, so they are in the same linear span. Also $c_1,c_2$ are linearly independent: their fourth entries force the coefficient of $c_2$ to vanish, then their first entries force the other coefficient to vanish. Elementary row operations preserve dependence relations among columns, but generally change the column space itself, which is why the original columns must be used.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$, and the column space is the linear span of its columns in ${\mathbb{R}}^m$. Its null space consists of the input vectors sent to zero:The rank of a matrix isThe equality of row and column dimensions is the row-rank/column-rank theorem. Each displayed space is a linear subspace of the indicated ambient vector space.

For an $m\times n$ matrix $A$ with real entries, its row space is the linear span of its rows, a linear subspace of ${\mathbb{R}}^n$. Its column space is the linear span of its columns, a linear subspace of ${\mathbb{R}}^m$ and the image of a linear map $x\mapsto Ax$. Its null space isa linear subspace of ${\mathbb{R}}^n$. Finally,Thus the rank of a matrix counts the number of linearly independent columns, equivalently rows. The nullity is $\dim \ker A=n-\mathrm{rank}A$ by the rank-nullity theorem.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$. The column space is the linear span of its columns in ${\mathbb{R}}^m$, equivalently $\{A\mathbf{x}:\mathbf{x}\in {\mathbb{R}}^n\}$. The null space isThe matrix rank is the dimension of the column space, which equals the dimension of the row space. In particular,The last equality is the rank-nullity theorem.

Interpret “distinct eigenvalues” as $n$ pairwise distinct eigenvalues in the chosen field. Select a nonzero eigenvector $z_i$ for each. A single nonzero vector is linearly independent. Suppose inductively that $z_1,\dots ,z_k$ are linearly independent. If adding $z_{k+1}$ made the set linearly dependent, its nontrivial relation would have a nonzero coefficient of $z_{k+1}$, since otherwise it would contradict the induction hypothesis. Solving for $z_{k+1}$ would place it in the linear span of the first $k$ eigenvectors.

Part (c) would then force ${\lambda }_{k+1}={\lambda }_i$ for some $i\le k$, contradicting pairwise distinctness. Mathematical induction therefore proves that all $n$ selected eigenvectors are linearly independent, forming a basis. If the field is $\mathbb{R}$, the assertion requires $n$ real eigenvalues; a real matrix can instead have nonreal eigenvalues and a eigenvector basis over the complex numbers.

Writing the columns as $a_1,a_2,a_3$, matrix multiplication givesThe column space is the linear span of these columns, so this particular linear combination belongs to it. The assertion holds for every vector of coefficients $y$.

Over the underlying field $F$, the row space is the linear span of the row vectors of $A$, a vector subspace of $F^n$. The column space is the linear span of its column vectors, a vector subspace of $F^m$. The null space isa vector subspace of the domain. The column space is the image of a linear map represented by $A$; the null space is its kernel.

For a Lie algebra representation of $\mathfrak{g}$ on $V$, let $\mathfrak{g}V$ denote the linear span of $xv$ for $x\in \mathfrak{g}$ and $v\in V$. The derived algebra of the semidirect product of a Lie algebra and a module is the displayed direct sum of vector spaces. Every Lie bracket has first component in $[\mathfrak{g},\mathfrak{g}]$ and second component in $\mathfrak{g}V$. Conversely, brackets of $(x,0)$ with $(y,0)$ span the former component, and brackets of $(x,0)$ with $(0,v)$ span the latter. Therefore this semidirect product of a Lie algebra and a module is a perfect Lie algebra exactly when $\mathfrak{g}$ is a perfect Lie algebra and $\mathfrak{g}V=V$. When $V\ne 0$, its nonzero abelian Lie algebra ideal prevents it from being a semisimple Lie algebra. The sl2 Lie algebra acting on its two-dimensional defining representation supplies a five-dimensional example.

In a torsion-free divisible Abelian group, division by a nonzero integer is unique. The rational scalar operations are therefore a definitional expansion. Formulas reduce to Boolean combinations of rational linear equations: eliminate an existential variable in a conjunction by solving any equation with nonzero coefficient; if none fixes the variable, the finitely many nontrivial disequalities exclude only finitely many choices in an infinite group. Equations with zero coefficient reduce to parameter conditions.

Thus over a finite parameter set a complete one-variable type either specifies an element of its linear span, or excludes every element of that span. An uncountable rational vector space realizes the latter type because a finite-parameter span is countable. It is consequently an omega-saturated model in the additive language.

A rank-one matrix update changes a matrix $A$ by an outer product $u{v}^T$. For real vectors $u,v\in {\mathbb{R}}^n$, its action on an input vector $x$ isThe dot product $v^{T}x$ reads one scalar from the input, and the correction moves the output along $u$. All correction outputs therefore lie in one linear span. If either vector is zero, the update is zero; otherwise the correction is a rank-one matrix. The full updated matrix need not have rank one.

For example, chooseThe update sends $x=(x_1,x_2{)}^T$ to the correction $(3x_1-x_2)(1,2{)}^T$. Although all four entries change, there is only one direction of correction. Adding it to the identity matrix gives $B=\left(\begin{array}{cc}4& -1\\ 6& -1\end{array}\right)$.

This special structure allows existing work to be reused. The matrix determinant lemma updates the determinant using one scalar, and the Sherman–Morrison formula updates the matrix inverse using an outer product. Such formulas are useful when a linear system changes slightly after a new observation or a parameter adjustment: the unchanged part can retain its LU decomposition. The term rank-one describes the rank of the correction, not the size of the entry changes. ^[ref].

For row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.

The representer theorem reduces regularized optimization over a reproducing-kernel Hilbert space to the span of kernel sections at the observed points.

The row space of a matrix is the linear span of its rows. It is the column space of the matrix transpose.

If the continuous dual space $X^{\prime }$ of a normed vector space is norm separable, then $X$ is separable. Choose a dense sequence $f_n$ in the dual unit ball and unit-ball points $x_n$ with $|f_n(x_n)|\ge \Vert f_n\Vert /2$. If their closed linear span were proper, the Hahn-Banach distance formula would give a norm-one functional $f$ vanishing on it. An $f_n$ within $1/4$ of $f$ would simultaneously satisfy $|f_n(x_n)|<1/4$ and $|f_n(x_n)|>3/8$, a contradiction. Rational linear combinations give a countable dense subset of $X$.

The converse fails by duality of l1 and l infinity: ${\ell }^1$ is separable, whereas ${\ell }^{\infty }$ contains uncountably many zero-one sequences at mutual distance one.

A subset $S$ of a vector space is a spanning set when its linear span is the whole vector space.

On the linear span of $1,x$, define $g(a1+bx)=a+b\lambda$. Affine spectral mapping gives $|a+b\lambda |\le \Vert a1+bx\Vert$, making this well-defined with norm one and $g(1)=1$. A norm-preserving Hahn-Banach theorem extension to the whole C-star algebra remains unital; the norm characterization of a positive functional on a C-star algebra makes it a state on a C-star algebra attaining the spectral value.

A sublattice is a subgroup of a Euclidean lattice. It is a Euclidean lattice in its real linear span, which may have smaller dimension than the original ambient Euclidean space. The equal-parity integer points in ${\mathbb{Z}}^d$ form a full-rank sublattice of index $2^{d-1}$: reduction of all coordinate differences modulo two gives the quotient.

For a family $\mathcal{F}$ of real-valued functions on a finite set and normalized inner product $⟨\cdot ,\cdot ⟩$, the test-function seminorm isIt measures the largest correlation of $g$ with an allowed test. It is a seminorm, and becomes a norm when the tests separate points.