CommeUnJeu · L1 MPSI

Affine subspaces

⌚ ~91 min ▢ 11 blocks ✓ 32 exercises ➣ Prerequisites : Vector spaces

Every element of a vector space $E$ can be read in two ways: as a vector (a displacement) or as a point (a location, once an origin is fixed). Under the second reading, the subspaces $F$ of $E$ are precisely the « flat sets through the origin »: lines through $0_E$, planes through $0_E$, and so on. The natural next step is to allow such flat sets to be translated away from the origin --- a line not through $0_E$, a plane that misses $0_E$. The resulting objects are called affine subspaces: sets of the form $\mathcal{F} = x + F$ where $F$ is a subspace and $x$ is any vector of $E$. The subspace $F$ is intrinsic to $\mathcal{F}$; it is called the direction of $\mathcal{F}$.
The chapter has two pedagogical goals. First, extend the geometric intuition of high-school affine geometry (lines, planes, parallels, intersections) to arbitrary vector spaces. Second, recognize, in several already-encountered problems --- linear systems, linear differential equations, arithmetic-geometric recurrences, polynomial interpolation --- the same recurring shape « set of solutions of $u(x) = a$ »: either empty, or an affine subspace of direction $\mathrm{Ker}\,u$. This unifying viewpoint is the headline result of the chapter.
Conventions used throughout. $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$; $E$ is a $\mathbb{K}$-vector space. Vector subspaces of $E$ are denoted with upright capitals $F, G, H, \ldots$; affine subspaces with calligraphic capitals $\mathcal{F}, \mathcal{G}, \mathcal{H}, \ldots$. The notation $B = A + \vec u$ is equivalent to $\overrightarrow{AB} = \vec u$.

I Affine structure: points$\virgule$ vectors$\virgule$ translation

I.1 Points and vectors in a vector space

A single element of $E$ can be read as either a point or a vector. The dictionary between the two readings is: pick an origin (the zero vector $0_E$); a point $M \in E$ then identifies with the displacement vector $\overrightarrow{OM} = M$. Two points $A, B$ define a displacement vector $\overrightarrow{AB} := B - A$ (zero if $A = B$); conversely a vector $\vec u$ added to a point $A$ produces a new point $B = A + \vec u$. The two readings differ only in vocabulary --- the underlying set is the same $E$.

Definition — Translation

For $\vec u \in E$, the translation by $\vec u$ is the map $$ t_{\vec u} : E \longrightarrow E, \qquad A \longmapsto A + \vec u. $$ The relation $B = A + \vec u$ is equivalent to $\overrightarrow{AB} = \vec u$.

Chasles relation

For three points $A, B, C \in E$, the Chasles relation reads $$ \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}. $$ Geometrically: « going from $A$ to $C$ directly » equals « going from $A$ to $B$ then from $B$ to $C$ ».

Example

In $\mathbb{R}^2$, take $A = (1, 2)$ and $B = (4, 6)$. Then $\overrightarrow{AB} = B - A = (3, 4)$, and the translation $t_{\overrightarrow{AB}}$ sends the origin $(0, 0)$ to $(3, 4)$ and the point $A$ to $A + \overrightarrow{AB} = B$.

Method — Reading an element as a point or a vector

To work in « point mode »: pick an origin (typically $0_E$). Each $M \in E$ is read as a point, and the vector $\overrightarrow{OM} = M$ is the displacement from the origin to $M$. To work in « vector mode », forget the origin and read each element as a free displacement. The two readings are interchangeable; the choice depends on whether you want to emphasize positions (affine objects: lines, planes) or displacements (vector operations: addition, scalar multiplication).

Skills to practice

Manipulating points$\virgule$ vectors and translations

I.2 Affine subspaces $\virgule$ definition and direction

A subspace $F$ of $E$ passing through $0_E$ is « anchored at the origin ». If we translate it by some vector $\vec u$, we obtain the set $\vec u + F$ --- still « flat » in $E$ (a line, a plane, \ldots), no longer containing $0_E$ unless $\vec u \in F$. Such translated subspaces are called affine subspaces of $E$; the original subspace $F$ is intrinsic to the translated set and is called its direction.

Definition — Affine subspace$\virgule$ direction

A subset $\mathcal{F} \subset E$ is an affine subspace of $E$ if there exist $x \in E$ and a vector subspace $F$ of $E$ such that $$ \mathcal{F} = x + F := \{x + f : f \in F\}. $$ The subspace $F$ is unique (see Proposition below) and called the direction of $\mathcal{F}$. Its elements are called direction vectors of $\mathcal{F}$.

Warning --- affine subspaces are non-empty

Since $\mathcal{F} = x + F$ contains $x$ (because $0_E \in F$), an affine subspace is always non-empty. The empty set is therefore not an affine subspace. This warning matters for the Theorem on linear equations $u(x) = a$ below, whose solution set is « either empty, or an affine subspace » --- the empty case is genuinely distinct.

Proposition — Uniqueness of the direction

Let $\mathcal{F}$ be an affine subspace of $E$. If $\mathcal{F} = x + F = x' + F'$ for some $x, x' \in E$ and subspaces $F, F'$ of $E$, then $F = F'$.

Proof

By symmetry, it suffices to show $F \subset F'$. Let $f \in F$.

Apply $\mathcal{F} = x + F = x' + F'$ at the element $x \in x + F$: $x = x' + f_1'$ for some $f_1' \in F'$.
Apply the same equality at $x + f \in x + F$: $x + f = x' + f_2'$ for some $f_2' \in F'$.

Subtracting: $f = (x + f) - x = (x' + f_2') - (x' + f_1') = f_2' - f_1' \in F'$ (since $F'$ is a subspace). Hence $F \subset F'$. By the same argument with the roles of $(x, F)$ and $(x', F')$ exchanged, one also obtains $F' \subset F$. Therefore $F = F'$.

Example

Every subspace $F$ of $E$ is an affine subspace: take $x = 0_E$, then $F = 0_E + F$, so $\mathcal{F} = F$ has direction $F$ itself. (The converse --- an affine subspace is vector if and only if it contains $0_E$ --- is recorded as a corollary right after the characterisation theorem of the next section.)

Example

In $\mathbb{R}^3$, the set $\mathcal{F} = \{(x, y, z) \in \mathbb{R}^3 : x + y + z = 1\}$ is an affine subspace of $\mathbb{R}^3$.
Verification. The point $(1, 0, 0)$ satisfies $x + y + z = 1$. For $(x, y, z) \in \mathbb{R}^3$, $(x, y, z) \in \mathcal{F} \iff (x - 1, y, z) \in F$ where $F = \{(\alpha, \beta, \gamma) : \alpha + \beta + \gamma = 0\}$. So $\mathcal{F} = (1, 0, 0) + F$, with direction $F$ (a vector plane of $\mathbb{R}^3$, dimension $2$).

Example

In $\mathbb{R}[X]$, $\mathcal{E} = \{P \in \mathbb{R}[X] : XP' + P = 2X\}$. We show that $\mathcal{E}$ is an affine subspace and identify its direction.
Particular solution. $P = X$ satisfies $X \cdot 1 + X = 2X$.
Reduction. For any $P$, $P \in \mathcal{E} \iff X(P - X)' + (P - X) = 0 \iff P - X \in H$ where $H = \{P \in \mathbb{R}[X] : XP' + P = 0\}$ (a fresh letter, since $E$ already denotes the ambient vector space). Hence $\mathcal{E} = X + H$, an affine subspace of $\mathbb{R}[X]$ with direction $H$.
Clarification on $H$. Observe that $XP' + P = (XP)'$, so $XP' + P = 0 \iff (XP)' = 0 \iff XP$ is a constant. Since $XP$ vanishes at $0$, the only constant value possible is $0$, i.e. $XP = 0$, i.e. $P = 0$. Therefore $H = \{0\}$ and $\mathcal{E} = \{X\}$ is reduced to a single point --- pedagogically clean: an affine subspace can reduce to a singleton (dimension $0$).

Skills to practice

Recognizing an affine subspace
Identifying the direction

I.3 Dimension of an affine subspace

The dimension of an affine subspace inherits from its direction. This makes the geometric naming of affine subspaces by their direction (line, plane, hyperplane) unambiguous.

Definition — Dimension of an affine subspace

Let $\mathcal{F}$ be an affine subspace of $E$ of direction $F$. When $F$ is of finite dimension, the dimension of $\mathcal{F}$ is $$ \dim \mathcal{F} := \dim F. $$

Example

A point (singleton $\{A\}$) is an affine subspace of dimension $0$ (direction $\{0_E\}$, a subspace of dimension $0$).
The whole space $E$ is an affine subspace of dimension $\dim E$ (direction $E$).
The affine subspace $\mathcal{E} = \{X\}$ of the previous example has dimension $0$.

Skills to practice

Computing the dimension of an affine subspace

I.4 Affine line$\virgule$ affine plane$\virgule$ affine hyperplane

Affine subspaces inherit naming from their direction. The vocabulary --- affine line, affine plane, affine hyperplane --- mirrors the vocabulary already established for the vector versions in the chapter Linear maps.

Definition — Affine line$\virgule$ affine plane$\virgule$ affine hyperplane

Let $\mathcal{F}$ be an affine subspace of $E$ of direction $F$.

If $\dim F = 1$, $\mathcal{F}$ is called an affine line.
If $\dim F = 2$, $\mathcal{F}$ is called an affine plane.
If $F$ is a hyperplane of $E$ (i.e. $F = \mathrm{Ker}\,\varphi$ for a non-zero linear form $\varphi$, or equivalently if $\dim F = \dim E - 1$ in finite dimension), $\mathcal{F}$ is called an affine hyperplane.

Example

Affine line in $\mathbb{R}^3$. The set $\{(1, 2, 3) + t(1, 0, -1) : t \in \mathbb{R}\}$ is an affine line of $\mathbb{R}^3$, passing through $(1, 2, 3)$ with direction $\mathrm{Vect}((1, 0, -1))$ (a vector line).
Affine plane in $\mathbb{R}^3$. The set $\{(0, 0, 1) + s(1, 0, 0) + t(0, 1, 0) : s, t \in \mathbb{R}\}$ is an affine plane: direction $\mathrm{Vect}(e_1, e_2)$ (the vector plane $z = 0$), translated to $z = 1$.

Example

In $\mathbb{R}_3[X]$ (dimension $4$): the set $\mathcal{P} = \{P \in \mathbb{R}_3[X] : P(0) = 1\}$ is an affine hyperplane of $\mathbb{R}_3[X]$.
Verification. The constant polynomial $P_0 = 1$ satisfies $P_0(0) = 1$. The direction is $\{P : P(0) = 0\}$, the kernel of the non-zero linear form $\varphi : P \mapsto P(0)$. So $\dim \mathcal{P} = \dim \mathbb{R}_3[X] - 1 = 3$.

Skills to practice

Working with affine lines$\virgule$ planes$\virgule$ hyperplanes

II Characterisation of affine subspaces

II.1 Characterisation by a point and the direction

Once a single point $A$ of an affine subspace is known, and its direction $F$ is known, the entire affine subspace is determined: $\mathcal{F} = A + F$. This is the analogue of « one point + direction vector » characterising a line in $\mathbb{R}^2$. It also gives a clean equality criterion: two affine subspaces coincide if and only if they share a point and have the same direction.

Theorem — Characterisation by direction and a point

Let $\mathcal{F}$ be an affine subspace of $E$ of direction $F$. Then for every $A \in \mathcal{F}$, $$ \mathcal{F} = A + F. $$ In particular, two affine subspaces are equal if and only if they have the same direction and share a common point.

Proof

By definition, $\mathcal{F} = x + F$ for some $x \in E$. Given $A \in \mathcal{F}$, write $A = x + f_0$ with $f_0 \in F$. Then $x = A - f_0$, so $$ \mathcal{F} = x + F = (A - f_0) + F = A + (F - f_0). $$ Since $F$ is a vector subspace, the translation $u \mapsto u - f_0$ is a bijection of $F$ onto $F$ (inverse: $u \mapsto u + f_0$); hence $F - f_0 = F$. Therefore $\mathcal{F} = A + F$.
Equality criterion. If $\mathcal{F} = \mathcal{G}$, the two share every point, and by uniqueness of the direction (established earlier in this chapter) they have the same direction. Conversely, suppose $\mathcal{F}$ and $\mathcal{G}$ have the same direction $F$ and share a point $A$. By what was just proved, $\mathcal{F} = A + F$ and $\mathcal{G} = A + F$, hence $\mathcal{F} = \mathcal{G}$.

Corollary --- vector subspaces among affine subspaces

An affine subspace $\mathcal{F}$ of $E$ is a vector subspace of $E$ if and only if $0_E \in \mathcal{F}$. Indeed, if $0_E \in \mathcal{F}$, applying the theorem with $A = 0_E$ gives $\mathcal{F} = 0_E + F = F$, a vector subspace. Conversely, every vector subspace contains $0_E$.

Method — Checking that a point belongs to an affine subspace

To check whether $M \in E$ belongs to a given affine subspace $\mathcal{F} = A + F$: compute the displacement vector $M - A$ and verify it lies in the direction $F$. Equivalently, $M \in \mathcal{F} \iff M - A \in F$.

Example

Does $M = (4, -1, 0)$ belong to the affine line $\mathcal{D}$ of $\mathbb{R}^3$ passing through $A = (1, 2, 3)$ with direction $\mathrm{Vect}((1, -1, -1))$?
Compute $M - A = (3, -3, -3) = 3 \cdot (1, -1, -1)$. Since $(1, -1, -1) \in \mathrm{Vect}((1, -1, -1))$, also $3 \cdot (1, -1, -1) \in \mathrm{Vect}((1, -1, -1))$. So $M - A \in F$, hence $M \in \mathcal{D}$.

Skills to practice

Checking that a point belongs to an affine subspace

II.2 Parallelism

Two affine subspaces are parallel when their directions are comparable by inclusion. When they have the same finite dimension, this reduces to having the same direction. They are strictly parallel when they are parallel and disjoint --- the high-school picture of two parallel lines that never meet.

Definition — Parallelism

Let $\mathcal{F}$ and $\mathcal{G}$ be two affine subspaces of $E$ with respective directions $F$ and $G$.

$\mathcal{F}$ and $\mathcal{G}$ are said to be parallel when their directions are comparable by inclusion: $$ F \subset G \qquad \text{or} \qquad G \subset F. $$ When they have the same finite dimension, this is equivalent to $F = G$.
$\mathcal{F}$ and $\mathcal{G}$ are said to be strictly parallel when they are parallel and disjoint: $$ (F \subset G \ \text{or}\ G \subset F) \qquad \text{and} \qquad \mathcal{F} \cap \mathcal{G} = \emptyset. $$

Example

In $\mathbb{R}^3$: $\mathcal{D}_1 = \{(0, 0, 0) + t(1, 0, 0) : t \in \mathbb{R}\}$ and $\mathcal{D}_2 = \{(0, 0, 1) + t(1, 0, 0) : t \in \mathbb{R}\}$ are two distinct affine lines with the same direction $\mathrm{Vect}(e_1)$. They are parallel (same direction). Moreover the first lies in the plane $z = 0$ and the second in $z = 1$, so $\mathcal{D}_1 \cap \mathcal{D}_2 = \emptyset$: they are strictly parallel.

Skills to practice

Recognizing parallel affine subspaces

II.3 Intersection of affine subspaces

Unlike vector subspaces --- which always share $0_E$ and therefore always have non-empty intersection --- two affine subspaces can be disjoint. When their intersection is non-empty, it is itself an affine subspace, with direction the intersection of the individual directions.

Theorem — Intersection of affine subspaces

Let $(\mathcal{F}_i)_{i \in I}$ be a non-empty family of affine subspaces of $E$ with respective directions $(F_i)_{i \in I}$. Then either $$ \bigcap_{i \in I} \mathcal{F}_i = \emptyset, \qquad \text{or} \qquad \bigcap_{i \in I} \mathcal{F}_i \text{ is an affine subspace of direction } \bigcap_{i \in I} F_i. $$

Proof

Suppose $\bigcap \mathcal{F}_i \ne \emptyset$ and pick $A \in \bigcap \mathcal{F}_i$. By the previous Theorem, $\mathcal{F}_i = A + F_i$ for every $i$. Let $F := \bigcap_i F_i$. We show $\bigcap_i \mathcal{F}_i = A + F$ by double inclusion.

$(\subset)$ Let $M \in \bigcap \mathcal{F}_i$. For each $i$, $M \in \mathcal{F}_i = A + F_i$, so $M - A \in F_i$. Taking intersection over $i$: $M - A \in \bigcap_i F_i = F$. Hence $M \in A + F$.
$(\supset)$ For every $i$, $F \subset F_i$, so $A + F \subset A + F_i = \mathcal{F}_i$. Intersecting over $i$: $A + F \subset \bigcap_i \mathcal{F}_i$.

Method — Computing an intersection of affine subspaces

To compute the intersection $\mathcal{F} \cap \mathcal{G}$ given by Cartesian equations: stack the equations into a single system. If the system is compatible (admits a solution), the solution set is the intersection --- an affine subspace of $E$, of direction $F \cap G$. If the system is incompatible, the intersection is empty.

Example

In $\mathbb{R}^3$, compute the intersection of the two affine planes $$ \mathcal{P}_1 : x + y - z = 1, \qquad \mathcal{P}_2 : 2x - y + z = 0. $$ Adding the two equations: $3x = 1$, so $x = \tfrac{1}{3}$. Substituting back into the first: $\tfrac{1}{3} + y - z = 1$, i.e. $y - z = \tfrac{2}{3}$. Setting $z = s$ as parameter: $y = \tfrac{2}{3} + s$. Solution set: $$ \mathcal{P}_1 \cap \mathcal{P}_2 = \left\{ \left(\tfrac{1}{3},\ \tfrac{2}{3} + s,\ s\right) : s \in \mathbb{R} \right\} = \left(\tfrac{1}{3}, \tfrac{2}{3}, 0\right) + s \cdot (0, 1, 1), $$ an affine line of direction $\mathrm{Vect}((0, 1, 1))$.

Example

In $\mathbb{R}^3$, the two parallel planes $\mathcal{P}_1 : z = 0$ and $\mathcal{P}_2 : z = 1$ have empty intersection: the system $\{z = 0, z = 1\}$ is incompatible. So $\mathcal{P}_1 \cap \mathcal{P}_2 = \emptyset$.

Skills to practice

Computing an intersection

III Linear equations and affine subspaces

III.1 Solutions of a linear equation form an affine subspace

The second goal of the chapter is now within reach. Many problems already encountered --- linear systems, linear differential equations, arithmetic-geometric recurrences, polynomial interpolation --- share the same shape: solve $u(x) = a$ where $u$ is a linear map between vector spaces and $a$ is a target vector. The set of solutions is either empty or an affine subspace, of direction the kernel of $u$. This unifying viewpoint reduces all those problems to the same two-step recipe: find a particular solution; describe the homogeneous solutions.

Theorem — Linear equation $u(x)$ $\equal$ $a$

Let $E$ and $G$ be two $\mathbb{K}$-vector spaces, let $u \in \mathcal{L}(E, G)$, and let $a \in G$. The solution set $$ \mathcal{S}_a := \{x \in E : u(x) = a\} $$ is either empty or an affine subspace of $E$ of direction $\mathrm{Ker}\,u$.

Proof

Suppose $\mathcal{S}_a \ne \emptyset$ and pick a particular solution $x_0 \in \mathcal{S}_a$ (i.e. $u(x_0) = a$). For any $x \in E$: $$ \begin{aligned} x \in \mathcal{S}_a &\iff u(x) = a && \text{(definition)} \\ &\iff u(x) - u(x_0) = 0 && \text{(subtract $a = u(x_0)$)} \\ &\iff u(x - x_0) = 0 && \text{(linearity of $u$)} \\ &\iff x - x_0 \in \mathrm{Ker}\,u. \end{aligned} $$ Hence $\mathcal{S}_a = x_0 + \mathrm{Ker}\,u$, an affine subspace of direction $\mathrm{Ker}\,u$.

Schematic --- solutions as particular plus kernel

The structure « solutions $=$ one particular solution $+$ kernel » can be pictured: the kernel $\mathrm{Ker}\,u$ is a vector subspace of $E$ (passing through the origin); the solution set $\mathcal{S}_a$ is the translated copy passing through any particular solution $x_0$.

Corollary --- level sets of a linear map

For $u \in \mathcal{L}(E, G)$ and $a \in G$, the preimage $u^{-1}(\{a\})$ is by definition the set $\{x \in E : u(x) = a\}$, i.e. $\mathcal{S}_a$. By the Theorem, $u^{-1}(\{a\})$ is either empty or an affine subspace of $E$ of direction $\mathrm{Ker}\,u$.
Special case --- affine hyperplane: if $\varphi$ is a non-zero linear form on $E$ and $\alpha \in \mathbb{K}$, then $$ \{x \in E : \varphi(x) = \alpha\} $$ is an affine hyperplane of $E$, of direction $\mathrm{Ker}\,\varphi$ which is a vector hyperplane. It is automatically non-empty: since $\varphi \ne 0$ there exists $x_1 \in E$ with $\varphi(x_1) \ne 0$, and then $x_0 = \tfrac{\alpha}{\varphi(x_1)} x_1$ satisfies $\varphi(x_0) = \alpha$. This reframes the earlier examples $$ x + y + z = 1 \quad \text{in } \mathbb{R}^3 $$ and $$ P(0) = 1 \quad \text{in } \mathbb{R}_3[X] $$ as level sets of explicit linear forms.

Method — Structure-of-solutions algorithm

To solve a linear equation $u(x) = a$:

Find a particular solution $x_0 \in E$ satisfying $u(x_0) = a$ (any single solution will do).
Determine $\mathrm{Ker}\,u$, the set of solutions of the associated homogeneous equation $u(h) = 0$.
The full solution set is $\mathcal{S}_a = x_0 + \mathrm{Ker}\,u = \{x_0 + h : h \in \mathrm{Ker}\,u\}$.

If no particular solution exists, $\mathcal{S}_a = \emptyset$.

Skills to practice

Applying the structure-of-solutions theorem

III.2 Application 1: linear systems

A linear system $AX = B$ with $A \in \mathcal{M}_{n,p}(\mathbb{K})$ and $B \in \mathbb{K}^n$ is exactly the equation $u(X) = B$ where $u : \mathbb{K}^p \to \mathbb{K}^n$, $X \mapsto AX$ (the linear map canonically associated to $A$). The theorem on linear equations applies directly.

Example

Solve the system $\{x + y - z = 1,\ 2x - y + z = 0\}$ in $\mathbb{R}^3$ via the structure-of-solutions algorithm.

Particular solution. Set $z = 0$: the system becomes $\{x + y = 1, 2x - y = 0\}$. Adding: $3x = 1$, $x = \tfrac{1}{3}$; then $y = \tfrac{2}{3}$. So $x_0 = (\tfrac{1}{3}, \tfrac{2}{3}, 0)$.
Homogeneous kernel. Solve $\{x + y - z = 0,\ 2x - y + z = 0\}$. Adding: $3x = 0$, $x = 0$; then $y = z$. Setting $z = s$: $\mathrm{Ker}\,u = \{(0, s, s) : s \in \mathbb{R}\} = \mathrm{Vect}((0, 1, 1))$.
Full solution. $\mathcal{S} = (\tfrac{1}{3}, \tfrac{2}{3}, 0) + s \cdot (0, 1, 1)$ for $s \in \mathbb{R}$.

Cross-reference. This is exactly the affine line found earlier by direct intersection in the « intersection of affine subspaces » section; the structure-of-solutions algorithm gives the same answer through the linear-equation viewpoint.

Skills to practice

Describing the solutions of a linear system as an affine subspace

III.3 Application 2: classical linear problems (overview)

The same structure-of-solutions theorem appears in three other classical settings already studied or to be studied. The detailed derivations live in the corresponding chapters; here we only recognize the pattern.

Example — Linear ODE of order 1

On an interval $I \subset \mathbb{R}$, for continuous functions $a, b \in \mathcal{C}^0(I, \mathbb{K})$, the linear ODE $$ y'(x) + a(x) y(x) = b(x) $$ is the equation $u(y) = b$ where $u : \mathcal{C}^1(I, \mathbb{K}) \to \mathcal{C}^0(I, \mathbb{K})$, $y \mapsto y' + a y$ is linear. The general theorem on linear equations says that the solution set is empty or an affine subspace of $\mathcal{C}^1(I, \mathbb{K})$ of direction $\mathrm{Ker}\,u$ (the homogeneous solutions); the existence theorem from the chapter Differential equations ensures that, in this setting (continuous $a$ and $b$ on an interval), it is in fact non-empty. The dimension of $\mathrm{Ker}\,u$ and the explicit form of solutions are derived there.

Example — Linear ODE of order 2$\virgule$ constant coefficients

On an interval $I \subset \mathbb{R}$, for $p, q \in \mathbb{K}$ and $f \in \mathcal{C}^0(I, \mathbb{K})$, the linear ODE $$ y''(x) + p y'(x) + q y(x) = f(x) $$ is $u(y) = f$ where $u : \mathcal{C}^2(I, \mathbb{K}) \to \mathcal{C}^0(I, \mathbb{K})$, $y \mapsto y'' + p y' + q y$ is linear. The general theorem gives « empty or affine of direction $\mathrm{Ker}\,u$ »; the existence theorem from Differential equations ensures non-emptiness in this setting. Admitted in the constant-coefficient case: $\mathrm{Ker}\,u$ has dimension $2$.

Example — Arithmetic-geometric recurrence

For $a, b \in \mathbb{K}$, consider the recurrence $u_{n+1} = a u_n + b$. View as $\Phi(u) = b \cdot \mathbf{1}$ where $\Phi : \mathbb{K}^\mathbb{N} \to \mathbb{K}^\mathbb{N}$, $(u_n) \mapsto (u_{n+1} - a u_n)$, and $\mathbf{1} = (1, 1, 1, \ldots)$. Two cases.

Case $a \ne 1$: choose the constant particular solution $\ell = b / (1 - a)$. The homogeneous solutions are geometric: $u_n = C a^n$ (with the usual convention $a^0 = 1$, including for $a = 0$). So $u_n = \ell + C a^n$.
Case $a = 1$: the recurrence becomes $u_{n+1} = u_n + b$. Choose the particular solution $u_n = n b$. The homogeneous solutions are the constant sequences. So $u_n = n b + C$.

In both cases, the solution set is « one particular sequence + 1-dimensional kernel » --- an affine subspace of $\mathbb{K}^\mathbb{N}$.

Example — Polynomial interpolation

Given $n + 1$ distinct real numbers $a_0, \ldots, a_n$ and target values $b_0, \ldots, b_n \in \mathbb{R}$, find $P \in \mathbb{R}_n[X]$ with $P(a_i) = b_i$ for every $i$. This is $\Phi(P) = (b_0, \ldots, b_n)$ where $\Phi : \mathbb{R}_n[X] \to \mathbb{R}^{n+1}$, $P \mapsto (P(a_0), \ldots, P(a_n))$. By an exercise of Linear maps, $\Phi$ is a bijection (since the $a_i$ are distinct, $\mathrm{Ker}\,\Phi = \{0\}$). So $\mathcal{S}$ has direction $\{0\}$ and is reduced to a single point: the unique Lagrange interpolation polynomial.

Method — Recognising the linear-equation pattern

Whenever an equation reads $u(x) = a$ with $u$ a linear map between two vector spaces and $a$ a target in the codomain, the structure-of-solutions theorem applies: the solution set is empty or an affine subspace of direction $\mathrm{Ker}\,u$. The structure-of-solutions algorithm then reduces the problem to two ingredients:

one particular solution;
the kernel of $u$.

Skills to practice

Recognizing the linear-equation pattern