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A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

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]

For a fixed formal pseudodifferential operator $P={\sum }_j{p}_j{\partial }^j$ and a monic differential operator $L={\partial }^m+{\sum }_{l=0}^{m-1}{u}_l{\partial }^l$, cyclicity of the Adler trace gives $l_P(L)=\mathrm{Tr}(PL)$ with $c=\int p_{-m-1}dx$ and $a_l=\mathrm{res}({\partial }^{l}P)={\sum }_{k=0}^l\left(\genfrac{}{}{0ex}{}{l}{k}\right)p_{k-l-1}^{(k)}$. The function is affine in the variable coefficients; its variational derivative with respect to $u_l$ is $a_l$.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

A Hilbert space is a real or complex inner product space that is complete in the norm $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Completeness means that every Cauchy sequence has a limit in the space: if its terms become arbitrarily close in this norm, they converge to an allowed vector. The inner product supplies lengths and orthogonality; completeness ensures that limiting approximation procedures do not leave the space.

Finite-dimensional Euclidean spaces are examples, but the same geometry also works for sequences and functions. In the l2 sequence space,In the L2 space on $[-1,1]$ with Lebesgue measure,These formulas use the convention linear in the first argument. Elements of the L2 space are equivalence classes of square-integrable functions agreeing almost everywhere, so zero distance really identifies the same element. A vector can therefore be a whole function, and its squared length can measure an integrated squared error.

For example, approximating $f(x)=x^2$ by $a+bx$ is a nearest-point problem in this Hilbert space. The allowed approximations form the linear span of $1,x$, and the orthogonal projection is the constant $1/3$. Its residual $x^2-1/3$ is orthogonal to both allowed directions because its integral and its inner product with $x$ vanish. The geometry therefore identifies the best approximation without searching through every pair of coefficients.

This is useful whenever a complicated vector must be represented by a tractable collection of directions. Fourier series approximate functions with orthogonal oscillations; the Parseval identity relates the sum of squared coefficients to the squared norm. Least-squares solutions approximate attainable data using a linear operator. The Hilbert projection theorem explains why a nearest element exists and is unique for a nonempty closed convex set. Distinct application settings share the same geometry of useful components and orthogonal residuals. Casey Rodriguez's MIT lectures develop these connections between Hilbert spaces, Fourier series, and nearest-point problems.^[ref]

A historical mathematical setting is David Hilbert's Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen (1912). Its preface states that it reprints six communications published during 1904–1910 on linear integral equations. This provides a concrete source for the integral-equation setting associated with the name, while the abstract definition above incorporates later developments.^[ref]

François Viète studied the connection between positive roots of a polynomial and the coefficients of the powers of the unknown. His 1591 presentation of algebraic notation used letters for both known and unknown quantities, with consonants for known quantities and vowels for unknown quantities. That move makes it possible to express a relation for whole families of polynomials, rather than only a numerical instance. His historical setting concerned positive roots and expressions satisfying dimensional homogeneity; the modern splitting field formulation above is a broader statement.^[ref]

Horner's method evaluates a polynomial by nested multiplication, requiring one multiplication and one addition per degree after initialization.

For $p(X)=a_n{X}^n+\cdots +a_0$, start with $b_n=a_n$ and compute $b_j=x{b}_{j+1}+a_j$ for $j=n-1,\dots ,0$. The final value is $b_0=p(x)$. Equivalently,This reorganization reuses intermediate results, avoiding separate computation of every power. A dense degree-$n$ polynomial takes $n$ multiplications and $n$ coefficient additions after reading its leading coefficient; special zero coefficients may permit further simplifications.

For our cubic, the coefficient list in descending order is $1,-3,0,4$. The zero is important: it occupies the missing $X$ term's position. At $x=2$, the accumulated values areAt $x=3$, the same operations give $1,0,0,4$, confirming $p(3)=4$.

The intermediate values do more than evaluate. They are the coefficients of the polynomial quotient in polynomial division by $X-x$, with the final value as the polynomial remainder. At $x=2$, the first three values give $X^2-X-2$, soAt $x=3$, they give $X^2$, so $p(X)=(X-3)X^2+4$. This links an efficient numerical calculation with the exact factor theorem.

William George Horner's 1819 paper concerns solving numerical equations by successive approximation. The Royal Society archive records the manuscript and its reading on 1 July 1819; it provides a historical reference for the method's name. ^[ref]

// Target: geometry-and-topology.bigb

An inner product assigns a scalar $⟨u,v⟩$ to two vectors and supplies their lengths and orthogonality. Using the convention linear in the first argument, it satisfiesThe resulting norm is $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Over the real numbers, conjugation has no effect and the form is symmetric and bilinear. Over the complex numbers, it is sesquilinear; the convention linear in the second argument is also common, so formulas for coefficients must match the stated convention.

The Euclidean example is $⟨u,v⟩={\sum }_i{u}_i\overline{v_i}$. For real $u=(2,1)$ and $v=(1,3)$ this gives $⟨u,v⟩=5$, while $\Vert u{\Vert }^2=5$ and $\Vert v{\Vert }^2=10$. If $r=(-1,2)$, then $⟨r,u⟩=0$: perpendicularity becomes an algebraic equation. The Cauchy-Schwarz inequality gives $|⟨u,v⟩|\le \Vert u\Vert \Vert v\Vert$, and in a real inner product space it makes the angle formula $\cos \theta =⟨u,v⟩/(\Vert u\Vert \Vert v\Vert )$ meaningful for nonzero vectors.

Different inner products express different measurements. The weighted example $⟨u,v⟩=4u_1{v}_1+u_2{v}_2$ on ${\mathbb{R}}^2$ gives $\Vert (x,y){\Vert }^2=4x^2+y^2$, so its unit-length vectors lie on an ellipse. For functions, an integral such as $⟨f,g⟩=\int f\overline{g}$ supplies the geometry of the L2 space. This lets orthogonal projection and least-squares solutions apply to curves as well as coordinate lists.

The uniform approximation closure of the integer-coefficient polynomials on $[0,1]$ consists precisely of the continuous functions taking integer values at both endpoints. Necessity follows because evaluations at zero and one are integers and convergent integer sequences are eventually constant. For sufficiency, the rounded Bernstein polynomial has integer monomial coefficients and differs uniformly by a vanishing amount from the Bernstein polynomial, which converges uniformly to the target.

Every $R\in [0,1]$ occurs as the radius of convergence of a power series all of whose coefficients are positive integers. For $0<R\le 1$, use the ceiling function to choose $a_n=\lceil R^{-n}\rceil$. ThenFor $|z|<R$, the comparison test for series with a geometric series proves absolute convergence. For $|z|\ge R$, the bound $|a_n{z}^n|\ge (|z|/R{)}^n\ge 1$ violates the fact that terms of a convergent series tend to zero. Hence the radius of convergence is exactly $R$.

For $R=0$, use $a_n=n!$: at every $z\ne 0$ the ratio of successive term absolute values is $(n+1)|z|$, eventually exceeding two. The factorial coefficients therefore give divergence away from zero.

A power series with every coefficient a positive integer cannot have radius of convergence greater than one: $a_n\ge 1$ makes convergence impossible at $z=1$. If zero coefficients are allowed and only finitely many are nonzero, a polynomial instead has infinite radius of convergence.

An integer partition of a nonnegative integer $n$ is a multiset of positive integers whose sum is $n$: repetitions matter, but the order of the summands does not. Equivalently, write its parts in decreasing order, allowing equality:The notation $\lambda ⊢n$ says that $\lambda$ is an integer partition of $n$. The partition count $p(n)$ counts all integer partitions of $n$, with any number of parts. Zero has one empty integer partition, so $p(0)=1$. The partition of an integer article develops the same objects in representation theory of the symmetric group.^[ref]

Computed example. The seven integer partitions of five are exactlyThus $p(5)=7$. Sorting each list of positive summands into decreasing order prevents repetitions in this enumeration. Grouping by largest part verifies completeness: largest parts $5,4,3,2,1$ contribute respectively $1,1,2,2,1$ possibilities. For largest part $3$, the remaining $2$ is either $2$ or $1+1$; for largest part $2$, the remaining $3$ uses only $2$ and $1$, giving $2+1$ or $1+1+1$.

An integer partition keeps sizes without named positions. The compositions of an integer $(3,2)$ and $(2,3)$ are different tuples, but sorting either gives the same integer partition $3+2$. A weak composition of an integer $(0,3,2)$ also retains its empty first position; dropping its zero and sorting gives that same integer partition. Conversely, a set partition records which distinct elements belong to each block. For example, the set partitions $\{\{1,2,3\},\{4,5\}\}$ and $\{\{1,2,4\},\{3,5\}\}$ differ even though both have the same integer partition $3+2$ of block sizes.

The Ferrers diagram makes an integer partition into a shape. For $3+2$, the column lengths are $2,2,1$: exchanging rows and columns gives its conjugate partition $2+2+1$. Exchanging again recovers $3+2$, so this operation is a bijection. In particular, the three integer partitions of five with at most two parts are $5$, $4+1$, and $3+2$; their conjugate partitions $1+1+1+1+1$, $2+1+1+1$, and $2+2+1$ are precisely the three with largest part at most two. This visible symmetry explains the equal counts for these two restrictions. The square-cell version is the existing Young diagram.

Counting by a generating function. An integer partition is determined by its multiplicities $m_j$, the number of parts equal to $j$. Its size condition is ${\sum }_{j\ge 1}j{m}_j=n$. Consequently its ordinary generating function isProof outline. Choose the term $q^{j{m}_j}$ from the $j$th geometric series. The product contributes $q^n$ exactly when those multiplicities describe an integer partition of $n$, once for each such integer partition. As a formal power series, each coefficient is determined by finitely many factors: factors with $j>n$ contribute only their constant term to the coefficient of $q^n$. For $n=5$, multiplication through $j=5$ gives $1+q+2q^2+3q^3+5q^4+7q^5$ up to degree five, independently recovering $p(5)=7$. Restricting the factors gives the generating function for partitions with bounded parts. This is useful because restrictions on summands become choices of factors, while conjugate partitions turn restrictions on numbers of rows into restrictions on row lengths.^[ref]

For a bipartite elliptope matrix $X$, put $t=\log (1+\sqrt{2})$ and apply $\sinh (tx)$ within the two diagonal blocks and $\sin (tx)$ across them. Matching absolute power series coefficients give a positive semidefinite matrix by coefficient-dominated entrywise positivity, while $\sinh t=1$ gives unit diagonal. Gaussian hyperplane rounding then turns each cross-block correlation coefficient into $(2/\pi )\arcsin (\sin (t{X}_{ij}))=(2t/\pi )X_{ij}$ because $t<\pi /2$. The zero diagonal blocks of the bipartite objective eliminate every other contribution.

Suppose an analytic function near $c$ is not identically zero there. Its Taylor series has a first nonzero coefficient $a_m$, andThe term $a_m(z-c{)}^m$ is its leading Taylor term. Dividing by $(z-c{)}^m$ gives a limit of $a_m\ne 0$ as $z\to c$, so this term determines the leading local magnitude and angular behaviour. The remainder uses big O notation; it is smaller by one power near the centre.

For a real-valued analytic function of a real variable, an even integer $m$ gives the same nearby sign on both sides of $c$, while an odd integer $m$ gives opposite signs. For example, $f(x)=x^2(1+x)$ has leading Taylor term $x^2$ and is positive on both sides of zero sufficiently nearby; $g(x)=x^3(1+x)$ has leading Taylor term $x^3$ and changes sign there. The finite factor $1+x$ stays positive for $|x|<1$. The ordinary constant term is the leading one when $f(c)\ne 0$.

Take the polynomial $f(x)=1+x+x^2+x^3$ on $[-1,1]$ with unit weight. Using the four Legendre polynomials above, direct integration givesFor example, $⟨f,P_2⟩=\frac{1}{2}{\int }_{-1}^1(3x^4+2x^2-1)dx=4/15$: the omitted odd terms have zero integral. Dividing by $h_2=2/5$ gives $c_2=2/3$. Substituting all four coefficients proves the exact identity

The successive orthogonal projections onto degrees at most $m$ are thereforeHere $\Vert g{\Vert }_2^2={\int }_{-1}^{1}g(x{)}^{2}dx$ is the squared L2 norm, without division by the interval length. Each row retains the previous Legendre polynomial coefficients and appends one new term. The ordinary power coefficients can change: adding $\frac{2}{3}{P}_2$ changes the constant from $4/3$ to $1$, and adding $\frac{2}{5}{P}_3$ changes the linear coefficient from $8/5$ to $1$. Independence belongs to the chosen orthogonal polynomial coordinates.

The full squared L2 norm is $\Vert f{\Vert }_2^2=192/35$, equal to $32/9+128/75+8/45+8/175$. The residual budgets are the omitted terms:For a direct check independent of summing the budgets, $f-q_2=x^3-3x/5$, and integrating its square gives $2/7-12/25+6/25=8/175$. This residual is orthogonal to $1,x,x^2$, exactly the conditions for the best quadratic least-squares solution. An orthogonal projection need not be a polynomial interpolant: for example, $q_2(1)=18/5$ while $f(1)=4$. The approximation allocates error across the interval according to the chosen weighted inner product.

Start from a Casimir function $H_0$ of the first Poisson structure. Whenever possible, choose $H_n$ whose first Hamiltonian vector field is minus the second Hamiltonian vector field of $H_{n-1}$. A sequence extending indefinitely is the coefficient sequence of a formal Casimir function for the Poisson pencil. Its functions pairwise commute for both Poisson brackets. Existence at each stage is an additional solvability requirement; compatibility alone does not guarantee it.

Let $V$ be a vector space over a specified field $F$, and let $v_1,\dots ,v_n\in V$. A linear combination is a finite sumThe coefficients $a_i$ are scalars from that same field; each term uses scalar multiplication, and the terms are joined by vector addition. For coefficients in the real numbers, negative values reverse arrows and fractional values scale them by fractional amounts. The collection of attainable vectors is the linear span. The column picture in Strang's MIT notes interprets solving a linear system as choosing coefficients to reach a target vector.^[ref]

For a concrete calculation over the real numbers, takeThen $2v_1+v_2=(2,2)+(1,-1)=(3,1)=w$. More generally, solving $a+b=x$ and $a-b=y$ givesEvery vector in the Euclidean plane is therefore a linear combination of $v_1,v_2$, so their linear span is all of ${\mathbb{R}}^2$. The target $(3,1)$ has coefficients $(2,1)$; the target $(0,2)$ has coefficients $(1,-1)$, since $(1,1)-(1,-1)=(0,2)$. The displayed formula also makes the coefficients unique. Consequently $(v_1,v_2)$ is an ordered basis, with coordinate vector $[(x,y){]}_{(v_1,v_2)}=((x+y)/2,(x-y)/2{)}^T$.

The payoff is a reusable language for synthesis: linear combinations build vectors from chosen ingredients, linear span decides which targets are reachable, and linear independence decides whether the coefficients are unique. The ingredients can also be polynomials or functions, as long as they belong to a vector space over the specified field.

A finite family of vectors in a vector space over a specified field $F$ is linearly dependent if it has a nontrivial linear relation: ${\sum }_i{a}_i{v}_i=0$ with $a_i\in F$ and at least one coefficient nonzero. One can then solve for a vector whose coefficient is nonzero as a linear combination of the others; removing that entry preserves the linear span. Repeated vectors in a list give linear dependence: for example $(v,v)$ has the linear relation $1v+(-1)v=0$. A list keeps both entries even though the set $\{v,v\}$ has only one element. Strang's notes illustrate redundant columns by expressing one as the sum of two others.^[ref]

Keep $v_1=(1,1)$ and $v_2=(1,-1)$, and add $v_3=(2,0)=v_1+v_2$. The three vectors still have linear span ${\mathbb{R}}^2$, but nowThe two coefficient triples $(2,1,0)$ and $(1,0,1)$ differ by the nontrivial linear relation $(-1,-1,1)$. In fact all linear combination representations of this target areTo check both completeness and correctness, use $c_1{v}_1+c_2{v}_2+c_3{v}_3=(c_1+c_3)v_1+(c_2+c_3)v_2$. Linear independence of $v_1,v_2$ forces $c_1+c_3=2$ and $c_2+c_3=1$, giving exactly the displayed family. Redundancy therefore preserves attainable targets while allowing their coefficients to move. For solving a linear system, these alternatives matter; for assigning unique coordinate vectors, one instead chooses a basis.

The following example shows why linear independence depends on the scalar field.

A scalar linear differential equation of order $n\ge 1$ has the formHere the coefficients $a_j$ and the forcing term $g$ are prescribed functions of the independent variable $t$. A classical solution of a differential equation is an $n$-times differentiable function satisfying the identity at every point of its interval. The unknown function $y$ and its derivatives enter to the first power, without products between them or coefficients depending on $y$. Equivalently, the differential operator $L$ is a linear map: for constant scalars $\alpha ,\beta$,The forcing term may be zero, giving a homogeneous linear differential equation, or nonzero, giving an inhomogeneous linear differential equation. Linearity of $L$ does not mean that every function graph of a solution of a differential equation is a straight line.

To use the usual order-$n$ initial value problem theory on an interval $I$, assume that $a_n(t)\ne 0$ throughout $I$ and that all $a_j$ and $g$ are continuous functions there. Dividing by $a_n$ normalizes the highest derivative; specifying $y(t_0),y^{\prime }(t_0),\dots ,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one solution of a differential equation throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in OpenStax's definition and standard form of linear equations; the corresponding system theorem is in MIT's existence and uniqueness notes for linear systems.

The coefficients need not be linear functions of $t$. For example, $t^2{y}^{\prime }+y=0$ is a linear differential equation on $t>0$, although $t^2$ is quadratic. Its solution of a differential equation with initial condition $y(1)=1$ is $y(t)=e^{1/t-1}$: its derivative is $y^{\prime }=-y/t^2$, so substitution gives $t^2{y}^{\prime }+y=0$, and $y(2)=e^{-1/2}\approx 0.60653$. The zero of the highest coefficient at $t=0$ explains the domain restriction. In contrast, $y^{\prime }+y^2=0$, $y{y}^{\prime }+y=0$ and $y^{\prime }+\sin y=0$ are nonlinear ordinary differential equations because of their dependence on the unknown function. Variable coefficients and nonlinear ordinary differential equations are different issues; see MIT's definitions and examples of first-order linear equations.

The practical benefit of linearity is that it separates the effect of the initial condition from the effect of the forcing term. Once one particular solution $y_p$ satisfies $L[y_p]=g$, every solution of a differential equation isHere $\ker L$ is the kernel of a linear map, consisting of solutions of a differential equation to the associated homogeneous linear differential equation. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed forcing term is an affine space obtained by translating $\ker L$. If $g$ is not identically zero, the zero function is absent, so this family is not a vector space under the usual operations on functions. This short argument explains why a general solution is a particular solution plus a homogeneous solution.

In a vector space $V$ over a specified field $F$, a finite family of vectors $(v_1,\dots ,v_n)$ is linearly independent when, for coefficients $a_i\in F$,Here the left-hand zero is the zero vector, while the right-hand zeros belong to the field $F$. Thus linear independence excludes every nontrivial linear relation. For an infinite family, require this condition for every finite subfamily.^[ref]

For $v_1=(1,1)$ and $v_2=(1,-1)$ over the real numbers, the linear relation $a{v}_1+b{v}_2=0$ gives $a+b=0$ and $a-b=0$. Adding and subtracting yields $2a=2b=0$, hence $a=b=0$. This verifies linear independence directly and explains why the linear combination giving $(3,1)$ has exactly the coefficients $(2,1)$.

A linearly independent family cannot contain the zero vector: a coefficient of $1$ on that entry and $0$ on every other entry would be a nontrivial linear relation. The empty family is linearly independent, because its empty sum is the zero vector and there are no coefficients that could be nonzero. Linear independence identifies a collection without redundant vectors; combined with a spanning set, it gives a basis. The uniqueness of linear combination coefficients makes its representations unambiguous.

Consider the vectors $1,i$ in the complex numbers, where $i$ is the imaginary unit. With scalar field $\mathbb{R}$, a linear relation $a\cdot 1+b\cdot i=0$ has $a,b\in \mathbb{R}$. Taking its real part and imaginary part gives $a=b=0$, so the pair is linearly independent. Every complex number $x+iy$ also has coefficients $(x,y)$ in the real numbers, so $1,i$ form a basis and the dimension of a vector space is two.

With scalar field $\mathbb{C}$, the same pair is linearly dependent, sinceuses the nonzero coefficients $(i,-1)$ from the complex numbers. The single vector $1$ is now a basis, because $z=z\cdot 1$ for every complex number $z$, giving dimension of a vector space one. For example $3+i=3\cdot 1+1\cdot i=(3+i)\cdot 1+0\cdot i$: the second representation uses a coefficient unavailable over the real numbers. Specifying the field is part of the definition of linear independence.