CommeUnJeu · L1 PCSI

Linear maps

⌚ ~187 min ▢ 23 blocks ✓ 63 exercises ➣ Prerequisites : Vector spaces, Maps

A linear map is the structure-preserving map between vector spaces. It is to vector spaces what group morphisms are to groups and ring morphisms are to rings: a function that commutes with the two operations of the structure --- addition and scalar multiplication. Linear maps are the « bridges » along which generating families, free families, bases, and direct sum decompositions are transported from one vector space to another.
In finite dimension, the rank theorem ties together two numbers attached to a linear map $f : E \to F$ --- the dimension of $\mathrm{Ker}\,f$ (how many degrees of freedom collapse) and the rank $\mathrm{rg}(f) = \dim \mathrm{Im}\,f$ (how many degrees of freedom land in $F$). Their sum equals $\dim E$: a conservation law of degrees of freedom that drives almost every dimension argument of the chapter.
The chapter closes with two geometric application classes: projectors and symmetries relative to a direct sum decomposition, and linear forms and hyperplanes, which describe codimension-one subspaces. Throughout, $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$ (this assumption is needed in the symmetry-characterisation proof, where we divide by $2$). The systematic study of the dual space and dual bases is out of program (« L'étude de la dualité est hors programme »); coordinate forms appear only as a practical tool for writing equations of hyperplanes.

I Linear maps: definition and structure

I.1 Definition of a linear map

The starting definition is short: a linear map is a function from a vector space to a vector space that respects linear combinations. Two operations to preserve --- addition and scalar multiplication --- give two axioms, which we will soon collapse into a single « one-scalar » condition.

Definition — Linear map

Let $E$ and $F$ be two $\mathbb{K}$-vector spaces. A map $f : E \to F$ is called linear if $$ \forall (x, y) \in E^2,\ \forall \lambda \in \mathbb{K} : \quad f(x + y) = f(x) + f(y) \quad \text{and} \quad f(\lambda x) = \lambda f(x). $$ The set of linear maps from $E$ to $F$ is denoted $\textcolor{colordef}{\mathcal{L}(E, F)}$. A linear map from $E$ to itself is called an endomorphism of $E$; the set of endomorphisms of $E$ is denoted $\textcolor{colordef}{\mathcal{L}(E)}$. A linear map $E \to \mathbb{K}$ is called a linear form on $E$.
Immediate consequence. For any $f \in \mathcal{L}(E, F)$, $f(0_E) = 0_F$: indeed $f(0_E) = f(0 \cdot 0_E) = 0 \cdot f(0_E) = 0_F$.

Proposition — Linearity in one scalar

Let $f : E \to F$ be a map between two $\mathbb{K}$-vector spaces. Then $f$ is linear if and only if $$ \forall (x, y) \in E^2,\ \forall \lambda \in \mathbb{K} : \quad f(\lambda x + y) = \lambda f(x) + f(y). $$

Proof

Direct sense. Suppose $f$ is linear. Then $f(\lambda x + y) = f(\lambda x) + f(y) = \lambda f(x) + f(y)$ by the two axioms.
Converse. Suppose $f$ satisfies the one-scalar identity. We derive the two original axioms in three steps.
- Step 1 --- $f(0_E) = 0_F$. Apply the identity with $\lambda = 1$ and $x = y = 0_E$: $f(0_E + 0_E) = f(0_E) + f(0_E)$, i.e. $f(0_E) = 2 f(0_E)$, hence $f(0_E) = 0_F$.
- Step 2 --- scalar multiplication. Apply the identity with $y = 0_E$ and any $x \in E$, $\lambda \in \mathbb{K}$: $f(\lambda x) = \lambda f(x) + f(0_E) = \lambda f(x)$.
- Step 3 --- addition. Apply the identity with $\lambda = 1$ and any $x, y \in E$: $f(x + y) = f(x) + f(y)$.

Method — Showing a map is linear

To prove $f : E \to F$ is linear: check the one-scalar identity $f(\lambda x + y) = \lambda f(x) + f(y)$ on generic $x, y \in E$ and $\lambda \in \mathbb{K}$. One verification replaces two.

Method — Showing a map is not linear

To prove $f$ is not linear: exhibit one failure --- either $f(0_E) \ne 0_F$, or a concrete pair $(x, y)$ with $f(x + y) \ne f(x) + f(y)$, or a concrete pair $(\lambda, x)$ with $f(\lambda x) \ne \lambda f(x)$.

Example

The identity $\mathrm{Id}_E$ is an endomorphism of $E$. More generally, for any $\lambda \in \mathbb{K}$, the homothety $\lambda \mathrm{Id}_E : x \mapsto \lambda x$ is an endomorphism of $E$. In dimension $1$, the endomorphisms of $E$ are exactly the homotheties.

Example

The map $f : \mathbb{R}^2 \to \mathbb{R}^3$ defined by $f(x, y) = (2x - y,\ 3x + 2y,\ x + y)$ is linear.
Verification. For $(x, y), (x', y') \in \mathbb{R}^2$ and $\lambda \in \mathbb{R}$, $$ \begin{aligned} f(\lambda (x, y) + (x', y')) &= f(\lambda x + x',\ \lambda y + y') \\ &= \bigl(2(\lambda x + x') - (\lambda y + y'),\ 3(\lambda x + x') + 2(\lambda y + y'),\ (\lambda x + x') + (\lambda y + y')\bigr) \\ &= \lambda (2x - y, 3x + 2y, x + y) + (2x' - y', 3x' + 2y', x' + y') \\ &= \lambda f(x, y) + f(x', y'). \end{aligned} $$

Example

The two maps $g, h : \mathbb{R}^2 \to \mathbb{R}$ defined by $g(x, y) = x + y + 1$ and $h(x, y) = x^2 + y^2$ are not linear.
For $g$: $g(0, 0) = 1 \ne 0$, so $g(0_{\mathbb{R}^2}) \ne 0_\mathbb{R}$.
For $h$: $h(2, 2) = 8$, but $2 h(1, 1) = 4$, so $h(2 \cdot (1, 1)) \ne 2 h(1, 1)$.

Example

For a fixed $a \in \mathbb{K}$, the evaluation map $\mathrm{ev}_a : \mathbb{K}[X] \to \mathbb{K}$, $P \mapsto P(a)$, is a linear form on $\mathbb{K}[X]$.
The derivation $D : \mathbb{K}[X] \to \mathbb{K}[X]$, $P \mapsto P'$, is an endomorphism of $\mathbb{K}[X]$.

Example

The integration map $\mathcal{C}([0, 1], \mathbb{R}) \to \mathbb{R}$, $f \mapsto \int_0^1 f$, is a linear form on $\mathcal{C}([0, 1], \mathbb{R})$ by linearity of the integral.
On an interval $I$, the derivation $\mathcal{C}^1(I, \mathbb{R}) \to \mathcal{C}(I, \mathbb{R})$, $f \mapsto f'$, is a linear map by linearity of the derivative.

Skills to practice

Recognizing a linear map
Verifying linearity directly

I.2 Reference linear maps and geometric examples

Linearity is best understood geometrically. The standard geometric transformations of $\mathbb{R}^2$ and $\mathbb{R}^3$ that fix the origin --- rotations, homotheties, symmetries, and projections along a fixed direction --- are linear maps. They are the geometric prototypes to keep in mind throughout the chapter.

Example — Rotation of $\mathbb{R}^2$

For $\theta \in \mathbb{R}$, the rotation of $\mathbb{R}^2$ by angle $\theta$ centered at the origin is the map $$ R_\theta : (x, y) \longmapsto (x \cos \theta - y \sin \theta,\ x \sin \theta + y \cos \theta). $$ It is linear: every coordinate of $R_\theta(x, y)$ is a linear combination of $x$ and $y$.

Example — Homothety

For $\lambda \in \mathbb{K}$, the homothety of $\mathbb{R}^2$ of ratio $\lambda$ centered at the origin is $$ h_\lambda : (x, y) \longmapsto (\lambda x,\ \lambda y) = \lambda \cdot (x, y). $$ It is linear --- as an instance of the homothety $\lambda \mathrm{Id}_{\mathbb{R}^2}$ from the previous subsection.

Example — Axial symmetry

The axial symmetry of $\mathbb{R}^2$ with respect to the $x$-axis is the map $$ s : (x, y) \longmapsto (x,\ -y). $$ It is linear. (A geometric figure for symmetry appears later in the chapter, in the section on Projectors and symmetries, in the three-dimensional setting.)

Example — Projection of $\mathbb{R}^3$ on a plane

The projection of $\mathbb{R}^3$ on the plane $z = 0$ along the axis $Oz$ is the map $$ p : (x, y, z) \longmapsto (x,\ y,\ 0). $$ It is linear: the third coordinate is dropped, the first two are kept.

Method — Recognising a geometric linear map

Geometric transformations that fix the origin --- rotations, homotheties, axial and central symmetries, projections in $\mathbb{R}^2$ and $\mathbb{R}^3$ --- are linear. Conversely, a transformation that does not fix the origin (translation, affine map with non-zero constant term) is not linear.

Skills to practice

Identifying reference linear maps

I.3 Operations on linear maps

Two operations are available on linear maps. First, addition and scalar multiplication of functions: they make $\mathcal{L}(E, F)$ a $\mathbb{K}$-vector space. Second, composition: when the spaces match up, the composite of two linear maps is again linear, and on a single space $E$ the composition turns $\mathcal{L}(E)$ into a set of endomorphisms one can compose at will, with the identity $\mathrm{Id}_E$ as neutral element. Composition behaves well with respect to addition and scalars, but it is not commutative in general. Finally, the inverse of a bijective linear map is automatically linear, so the bijective endomorphisms of $E$ form a distinguished set $\mathrm{GL}(E) \subset \mathcal{L}(E)$.

Proposition — $\mathcal{L}(E\virgule\ F)$ is a vector space

Let $E$ and $F$ be $\mathbb{K}$-vector spaces. Then $\mathcal{L}(E, F)$ is a $\mathbb{K}$-vector space: it is a subspace of $F^E$ (the space of all functions $E \to F$), with the zero map $x \mapsto 0_F$ as its zero element.

Proof

The zero map is linear: $0(\lambda x + y) = 0_F = \lambda \cdot 0_F + 0_F = \lambda \cdot 0(x) + 0(y)$, so $\mathcal{L}(E, F)$ is non-empty. Let $f, g \in \mathcal{L}(E, F)$ and $\mu \in \mathbb{K}$. For $x, y \in E$ and $\lambda \in \mathbb{K}$, $$ \begin{aligned} (\mu f + g)(\lambda x + y) &= \mu f(\lambda x + y) + g(\lambda x + y) && \text{(definition of $\mu f + g$)} \\ &= \mu (\lambda f(x) + f(y)) + (\lambda g(x) + g(y)) && \text{(linearity of $f$ and $g$)} \\ &= \lambda (\mu f(x) + g(x)) + (\mu f(y) + g(y)) && \text{(rearranging)} \\ &= \lambda (\mu f + g)(x) + (\mu f + g)(y). \end{aligned} $$ So $\mu f + g \in \mathcal{L}(E, F)$. Hence $\mathcal{L}(E, F)$ is a subspace of $F^E$.

Theorem — Composition of linear maps

Let $E, F, G$ be $\mathbb{K}$-vector spaces, $f \in \mathcal{L}(E, F)$ and $g \in \mathcal{L}(F, G)$. Then $g \circ f \in \mathcal{L}(E, G)$.
The composition is bilinear: for $f, f' \in \mathcal{L}(E, F)$, $g, g' \in \mathcal{L}(F, G)$, $\lambda \in \mathbb{K}$, $$ g \circ (f + f') = g \circ f + g \circ f', \quad (g + g') \circ f = g \circ f + g' \circ f, \quad g \circ (\lambda f) = \lambda (g \circ f) = (\lambda g) \circ f. $$

Proof

For $x, y \in E$ and $\lambda \in \mathbb{K}$, $$ (g \circ f)(\lambda x + y) = g(f(\lambda x + y)) = g(\lambda f(x) + f(y)) = \lambda g(f(x)) + g(f(y)) = \lambda (g \circ f)(x) + (g \circ f)(y). $$ Hence $g \circ f$ is linear. Bilinearity follows directly: $(g \circ (f + f'))(x) = g(f(x) + f'(x)) = g(f(x)) + g(f'(x))$ by linearity of $g$, and similarly for the other identities.

Definition — Isomorphism

A linear map $f \in \mathcal{L}(E, F)$ that is bijective is called an isomorphism. When $E = F$, a bijective endomorphism is called an automorphism. The set of automorphisms of $E$ is denoted $\textcolor{colordef}{\mathrm{GL}(E)}$, the linear group of $E$.

Example

The identity $\mathrm{Id}_E : E \to E$ is an automorphism of $E$ (so $\mathrm{Id}_E \in \mathrm{GL}(E)$); its inverse is $\mathrm{Id}_E$ itself. More generally, for $\lambda \in \mathbb{K}^*$, the homothety $\lambda \mathrm{Id}_E$ is an automorphism of $E$ with inverse $\frac{1}{\lambda} \mathrm{Id}_E$. The map $\mathbb{R}^2 \to \mathbb{R}^2$, $(x, y) \mapsto (y, x)$ (swap of coordinates) is an automorphism: it is linear, bijective, and equal to its own inverse.

Theorem — Inverse of an isomorphism is linear

Let $f \in \mathcal{L}(E, F)$ be an isomorphism. Then $f^{-1} : F \to E$ is also linear, hence $f^{-1} \in \mathcal{L}(F, E)$.

Proof

Let $y, y' \in F$ and $\lambda \in \mathbb{K}$. Set $x := f^{-1}(y)$ and $x' := f^{-1}(y')$, so that $y = f(x)$ and $y' = f(x')$. By linearity of $f$, $$ f(\lambda x + x') = \lambda f(x) + f(x') = \lambda y + y'. $$ Applying $f^{-1}$ to both sides: $\lambda x + x' = f^{-1}(\lambda y + y')$, i.e. $f^{-1}(\lambda y + y') = \lambda f^{-1}(y) + f^{-1}(y')$. Hence $f^{-1}$ is linear.

Theorem — Composition of endomorphisms

Let $E$ be a $\mathbb{K}$-vector space. On $\mathcal{L}(E)$, the composition $\circ$ has the following properties:

it is associative: $(f \circ g) \circ h = f \circ (g \circ h)$;
it is distributive over addition, on the left and on the right: $f \circ (g + h) = f \circ g + f \circ h$ and $(f + g) \circ h = f \circ h + g \circ h$;
it is compatible with scalars: $\lambda (f \circ g) = (\lambda f) \circ g = f \circ (\lambda g)$ for $\lambda \in \mathbb{K}$;
it admits $\mathrm{Id}_E$ as neutral element: $f \circ \mathrm{Id}_E = \mathrm{Id}_E \circ f = f$.

However, composition is not commutative in general: $f \circ g \neq g \circ f$ for most pairs $f, g \in \mathcal{L}(E)$.

Proof

Associativity, the neutral role of $\mathrm{Id}_E$, and the fact that the composite of two endomorphisms is again an endomorphism are the general properties of the composition of maps, already established. The two distributivities are one-line verifications using linearity: for $x \in E$, $$ (f \circ (g + h))(x) = f(g(x) + h(x)) = f(g(x)) + f(h(x)) = (f \circ g + f \circ h)(x) $$ by linearity of $f$ (left distributivity), and $$ ((f + g) \circ h)(x) = f(h(x)) + g(h(x)) = (f \circ h + g \circ h)(x) $$ directly from the definition of $f + g$ (right distributivity). The scalar compatibilities follow the same way from linearity. Non-commutativity is shown in the Example below.

Example — {Non-commutativity in $\mathcal{L}(\mathbb{R}[X])$}

On $\mathbb{R}[X]$, consider the derivative $D : P \mapsto P'$ and the multiplication by $X$, $M : P \mapsto X P$. Both are endomorphisms. Compute: $$ (D \circ M)(P) = D(X P) = P + X P', \qquad (M \circ D)(P) = M(P') = X P'. $$ The difference $(D \circ M)(P) - (M \circ D)(P) = P$ for every $P$, i.e. $D \circ M - M \circ D = \mathrm{Id}_{\mathbb{R}[X]}$. So $D$ and $M$ do not commute, and $\mathcal{L}(\mathbb{R}[X])$ is non-commutative.

Method — Computing powers and polynomials of an endomorphism

For $f \in \mathcal{L}(E)$ and $n \in \mathbb{N}^*$, write $f^n := f \circ f \circ \cdots \circ f$ ($n$ times), with $f^0 := \mathrm{Id}_E$. Since composition is associative and distributive over addition, one can manipulate expressions $P(f) = a_n f^n + \cdots + a_1 f + a_0 \mathrm{Id}_E$ (for $P = a_n X^n + \cdots + a_0 \in \mathbb{K}[X]$) by the usual rules of computation, treating $f$ like an indeterminate and $\mathrm{Id}_E$ like the constant $1$.
Binomial formula $(f + g)^n$: valid only when $f \circ g = g \circ f$. Same caveat for the factorisation of $f^n - g^n$ and the geometric sum: they require $f$ and $g$ to commute.

Skills to practice

Composing linear maps; computing in $\mathcal{L}(E)$

I.4 Image and kernel

To every linear map $f : E \to F$ are attached two natural subspaces: the image, in $F$, of everything $f$ produces; and the kernel, in $E$, of everything $f$ kills (sends to $0_F$). Their relation to injectivity and surjectivity is direct: the kernel detects injectivity (it is $\{0\}$ iff $f$ is injective), and the image detects surjectivity (it equals $F$ iff $f$ is surjective).

Definition — Image

Let $f \in \mathcal{L}(E, F)$. The image of a subspace $A$ of $E$ is the set $$ f(A) := \{f(a) : a \in A\} \subset F. $$ The image of $f$ is $\textcolor{colordef}{\mathrm{Im}\,f := f(E)} = \{f(x) : x \in E\}$.
Direct consequence. $f$ is surjective if and only if $\mathrm{Im}\,f = F$.

Definition — Preimage and kernel

Let $f \in \mathcal{L}(E, F)$. The preimage of a subset $B \subset F$ is the set $$ f^{-1}(B) := \{x \in E : f(x) \in B\}. $$ The notation $f^{-1}(B)$ does not require $f$ to be bijective: it is a set definition, not the value of an inverse function. The kernel of $f$ is the preimage of $\{0_F\}$: $$ \textcolor{colordef}{\mathrm{Ker}\,f := f^{-1}(\{0_F\}) = \{x \in E : f(x) = 0_F\}}. $$

Proposition — Image of a subspace is a subspace

Let $f \in \mathcal{L}(E, F)$ and let $A$ be a subspace of $E$. Then $f(A)$ is a subspace of $F$. In particular, $\mathrm{Im}\,f$ is a subspace of $F$.

Proof

$0_F = f(0_E) \in f(A)$ since $0_E \in A$, so $f(A)$ is non-empty. Let $y, y' \in f(A)$ and $\lambda \in \mathbb{K}$: write $y = f(a)$ and $y' = f(a')$ with $a, a' \in A$. Then $\lambda y + y' = \lambda f(a) + f(a') = f(\lambda a + a')$ by linearity of $f$, and $\lambda a + a' \in A$ since $A$ is a subspace. Hence $\lambda y + y' \in f(A)$, so $f(A)$ is a subspace of $F$.

Proposition — Preimage of a subspace is a subspace

Let $f \in \mathcal{L}(E, F)$ and let $B$ be a subspace of $F$. Then $f^{-1}(B)$ is a subspace of $E$. In particular, $\mathrm{Ker}\,f$ is a subspace of $E$.

Proof

$f(0_E) = 0_F \in B$, so $0_E \in f^{-1}(B)$, hence $f^{-1}(B)$ is non-empty. Let $x, x' \in f^{-1}(B)$ and $\lambda \in \mathbb{K}$: $f(\lambda x + x') = \lambda f(x) + f(x') \in B$ since $f(x), f(x') \in B$ and $B$ is a subspace. Hence $\lambda x + x' \in f^{-1}(B)$, so $f^{-1}(B)$ is a subspace of $E$.

Proposition — Image of $\mathrm{Vect}$

Let $f \in \mathcal{L}(E, F)$ and let $X$ be a non-empty part of $E$. Then $$ f(\mathrm{Vect}(X)) = \mathrm{Vect}(f(X)), $$ where $f(X) = \{f(x) : x \in X\}$. In particular, if $(e_i)_{i \in I}$ is a generating family of $E$, then $\mathrm{Im}\,f = \mathrm{Vect}\bigl((f(e_i))_{i \in I}\bigr)$.

Proof

Double inclusion.

$(\subset)$ Let $y \in f(\mathrm{Vect}(X))$: write $y = f(x)$ with $x \in \mathrm{Vect}(X)$, i.e. $x = \lambda_1 x_1 + \cdots + \lambda_r x_r$ for some $x_i \in X$ and $\lambda_i \in \mathbb{K}$. Then $y = f(x) = \lambda_1 f(x_1) + \cdots + \lambda_r f(x_r) \in \mathrm{Vect}(f(X))$.
$(\supset)$ Let $z \in \mathrm{Vect}(f(X))$: $z = \mu_1 f(x_1) + \cdots + \mu_r f(x_r) = f(\mu_1 x_1 + \cdots + \mu_r x_r)$ for some $x_i \in X$ and $\mu_i \in \mathbb{K}$. As $\mu_1 x_1 + \cdots + \mu_r x_r \in \mathrm{Vect}(X)$, we get $z \in f(\mathrm{Vect}(X))$.

Proposition — Kernel detects injectivity

Let $f \in \mathcal{L}(E, F)$. Then $$ f \text{ is \textcolor{colorprop}{injective}} \iff \mathrm{Ker}\,f = \{0_E\}. $$

Proof

$(\Rightarrow)$ Suppose $f$ injective. As $f(0_E) = 0_F$, the only element of $E$ sent to $0_F$ is $0_E$, so $\mathrm{Ker}\,f = \{0_E\}$.
$(\Leftarrow)$ Suppose $\mathrm{Ker}\,f = \{0_E\}$. Let $x, y \in E$ with $f(x) = f(y)$. Then $f(x - y) = f(x) - f(y) = 0_F$, so $x - y \in \mathrm{Ker}\,f = \{0_E\}$, hence $x = y$. So $f$ is injective.

Example — Kernel and image of $D : P \mapsto P'$

On $\mathbb{K}[X]$, the derivation $D : P \mapsto P'$ has kernel $\mathrm{Ker}\,D = \mathbb{K}$ (the constants: $P' = 0$ iff $P$ is constant) and image $\mathrm{Im}\,D = \mathbb{K}[X]$ (every polynomial admits a primitive obtained term by term: a primitive of $\sum a_k X^k$ is $\sum \frac{a_k}{k+1} X^{k+1}$). So $D$ is surjective but not injective.

Example — Kernel and image of an $\mathbb{R}^3 \to \mathbb{R}^2$ map

Let $f : \mathbb{R}^3 \to \mathbb{R}^2$, $(x, y, z) \mapsto (2x + y - z,\ x - y)$.
Kernel. Solve $f(x, y, z) = 0$: $\{2x + y - z = 0,\ x - y = 0\}$, which gives $y = x$ and $z = 3x$. So $$ \mathrm{Ker}\,f = \{(x, x, 3x) : x \in \mathbb{R}\} = \mathrm{Vect}\bigl((1, 1, 3)\bigr), \quad \dim \mathrm{Ker}\,f = 1. $$ Image. As $(e_1, e_2, e_3)$ is a basis of $\mathbb{R}^3$, $\mathrm{Im}\,f = \mathrm{Vect}\bigl(f(e_1), f(e_2), f(e_3)\bigr) = \mathrm{Vect}\bigl((2, 1), (1, -1), (-1, 0)\bigr)$. The two vectors $(2, 1)$ and $(1, -1)$ are not proportional, hence they are free in $\mathbb{R}^2$, so $\mathrm{Im}\,f = \mathbb{R}^2$. So $f$ is surjective.

Method — Computing image and kernel

To compute $\mathrm{Ker}\,f$: solve the linear system $f(x) = 0_F$ component by component; the solution space is the kernel.
To compute $\mathrm{Im}\,f$ when $E$ has a finite basis $(e_1, \ldots, e_n)$: by the image-of-$\mathrm{Vect}$ proposition, $\mathrm{Im}\,f = \mathrm{Vect}\bigl(f(e_1), \ldots, f(e_n)\bigr)$; then extract a basis of $\mathrm{Im}\,f$ from these $n$ images using the « base extraite » algorithm of Finite-dimensional vector spaces.

Skills to practice

Computing image and kernel
Using the kernel for injectivity

II Determination of a linear map

II.1 Characterisation by image of a basis

A linear map between two vector spaces is entirely determined by its values on a basis of the source: knowing $u(e_i)$ for every vector $e_i$ of a basis is enough to compute $u(x)$ for every $x \in E$. Conversely, prescribing the images of basis vectors freely defines a unique linear map. This « basis $\mapsto$ image of basis » duality is the most-used construction principle of the chapter.

Theorem — Linear map determined by image of a basis

Let $E, F$ be two $\mathbb{K}$-vector spaces, $(e_i)_{i \in I}$ a basis of $E$, and $(y_i)_{i \in I}$ any family of vectors of $F$ indexed by the same $I$. Then there exists a unique $u \in \mathcal{L}(E, F)$ such that $$ \forall i \in I : \quad u(e_i) = y_i. $$

Proof

Analysis--synthesis.

Analysis (uniqueness). If such a $u$ exists, then for $x \in E$ with unique decomposition $x = \sum_{i \in I} x_i e_i$ (finite sum since $(e_i)$ is a basis), the linearity of $u$ forces $$ u(x) = u\Bigl(\sum_i x_i e_i\Bigr) = \sum_i x_i u(e_i) = \sum_i x_i y_i. $$ So $u$ is fully determined by the family $(y_i)$, hence unique.
Synthesis (existence). Define $u : E \to F$ by $u(x) := \sum_i x_i y_i$ where $(x_i)$ are the coordinates of $x$ in $(e_i)$. For $x = \sum_i x_i e_i$ and $x' = \sum_i x'_i e_i$, uniqueness of coordinates gives $\lambda x + x' = \sum_i (\lambda x_i + x'_i) e_i$, so $$ u(\lambda x + x') = \sum_i (\lambda x_i + x'_i) y_i = \lambda \sum_i x_i y_i + \sum_i x'_i y_i = \lambda u(x) + u(x'). $$ Hence $u$ is linear. Finally $u(e_j) = \sum_i \delta_{ij} y_i = y_j$ for every $j$.

Method — Defining a linear map by prescribing it on a basis

To define a linear map $u \in \mathcal{L}(E, F)$ when $E$ has a basis $(e_i)$: prescribe the images $u(e_i) \in F$ freely (no constraint between them). The Theorem guarantees that a unique linear map $u$ extends these prescriptions to all of $E$.

Example

Define $u : \mathbb{R}^3 \to \mathbb{R}^2$ by $u(e_1) = (2, 3)$, $u(e_2) = (1, 0)$, $u(e_3) = (1, -1)$, where $(e_1, e_2, e_3)$ is the canonical basis of $\mathbb{R}^3$. Then for $(x, y, z) \in \mathbb{R}^3$, $$ u(x, y, z) = x \cdot (2, 3) + y \cdot (1, 0) + z \cdot (1, -1) = (2x + y + z,\ 3x - z). $$

Skills to practice

Defining a linear map by image of a basis

II.2 Characterisation by restrictions to a direct sum

Instead of prescribing values vector by vector on a basis, one can prescribe « pieces » of $u$ on each summand of a direct sum $E = E_1 \oplus \cdots \oplus E_p$. The same analysis--synthesis argument works, this time with the unique decomposition of every $x \in E$ as a sum of vectors from each $E_i$.

Theorem — Linear map determined by restrictions to a direct sum

Let $E = E_1 \oplus E_2 \oplus \cdots \oplus E_p$ be a direct sum decomposition, $F$ a $\mathbb{K}$-vector space, and $u_i \in \mathcal{L}(E_i, F)$ a linear map for each $i \in \llbracket 1, p \rrbracket$. Then there exists a unique $u \in \mathcal{L}(E, F)$ such that $$ \forall i \in \llbracket 1, p \rrbracket : \quad u_{|E_i} = u_i. $$

Proof

Analysis--synthesis on the unique decomposition $x = x_1 + \cdots + x_p$ with $x_i \in E_i$.

Analysis. If such a $u$ exists, then $u(x) = u(x_1 + \cdots + x_p) = u(x_1) + \cdots + u(x_p) = u_1(x_1) + \cdots + u_p(x_p)$, so $u$ is uniquely determined by the $u_i$.
Synthesis. Define $u(x) := u_1(x_1) + \cdots + u_p(x_p)$. The decomposition $x = x_1 + \cdots + x_p$ being unique, $u$ is well-defined. Linearity: for $x, x' \in E$ with $x = \sum x_i$, $x' = \sum x'_i$, and $\lambda \in \mathbb{K}$, $\lambda x + x' = \sum (\lambda x_i + x'_i)$ (the direct sum is preserved by linear combinations), so $u(\lambda x + x') = \sum u_i(\lambda x_i + x'_i) = \lambda \sum u_i(x_i) + \sum u_i(x'_i) = \lambda u(x) + u(x')$. Finally, for $x \in E_i$, its decomposition has only the $i$-th summand non-zero, so $u(x) = u_i(x)$.

Method — Defining a linear map by pieces on a direct sum

To define $u \in \mathcal{L}(E, F)$ when $E = F_1 \oplus F_2$: prescribe the restrictions $u_{|F_1}$ and $u_{|F_2}$ independently. The two pieces extend uniquely to a linear map on all of $E$. This is the standard construction for projectors, symmetries, and any map whose behaviour is described separately on two complementary subspaces.

Example

In $\mathbb{R}^3 = F \oplus G$ with $F = \{(x, y, 0) : x, y \in \mathbb{R}\}$ (the plane $z = 0$) and $G = \mathrm{Vect}\bigl((0, 0, 1)\bigr)$ (the $z$-axis), define $u \in \mathcal{L}(\mathbb{R}^3)$ by $u_{|F} = \mathrm{Id}_F$ and $u_{|G} = 0$. Then $u(x, y, z) = (x, y, 0)$: this is the projection on $F$ along $G$ (a projector --- see the section on Projectors and symmetries below).

Skills to practice

Defining a linear map by pieces on a direct sum

II.3 Isomorphisms and dimension

The image of a basis under a linear map captures injectivity, surjectivity, and bijectivity. In finite dimension this leads to the program-mandated statement that two spaces are isomorphic if and only if they have the same dimension: in particular, every finite-dimensional $\mathbb{K}$-vector space of dimension $n$ is isomorphic to $\mathbb{K}^n$.

Theorem — Linear map vs image of a basis

Let $E$ be a $\mathbb{K}$-vector space of finite dimension, $(e_1, \ldots, e_n)$ a basis of $E$, $F$ a $\mathbb{K}$-vector space (not assumed finite-dimensional), and $u \in \mathcal{L}(E, F)$. Set $y_i := u(e_i)$ for $i \in \llbracket 1, n \rrbracket$. Then:

$u$ is injective $\iff$ $(y_1, \ldots, y_n)$ is free in $F$;
$u$ is surjective $\iff$ $(y_1, \ldots, y_n)$ generates $F$;
$u$ is bijective $\iff$ $(y_1, \ldots, y_n)$ is a basis of $F$.

Proof

We use $\mathrm{Im}\,u = \mathrm{Vect}(y_1, \ldots, y_n)$ (Proposition « image of $\mathrm{Vect}$ ») and the « kernel detects injectivity » Proposition.

Surjective $\iff$ $(y_i)$ generates $F$. Direct from $\mathrm{Im}\,u = \mathrm{Vect}(y_1, \ldots, y_n)$: $u$ surjective $\iff$ $\mathrm{Im}\,u = F$ $\iff$ $(y_i)$ generates $F$.
Injective $\iff$ $(y_i)$ is free. For $\lambda_1, \ldots, \lambda_n \in \mathbb{K}$, $u(\sum \lambda_i e_i) = \sum \lambda_i y_i$. So $\mathrm{Ker}\,u = \{\sum \lambda_i e_i : \sum \lambda_i y_i = 0_F\}$. Hence $\mathrm{Ker}\,u = \{0_E\}$ (i.e. $u$ injective) iff $\sum \lambda_i y_i = 0_F$ implies all $\lambda_i = 0$, i.e. iff $(y_i)$ is free.
Bijective $\iff$ $(y_i)$ is a basis. Bijective = injective + surjective; basis = free + generating. Combine the two previous points.

Theorem — Isomorphism preserves dimension

Let $E$ be a $\mathbb{K}$-vector space of finite dimension and let $F$ be a $\mathbb{K}$-vector space.
(i) If $E$ and $F$ are isomorphic, then $F$ is finite-dimensional and $\dim F = \dim E$.
(ii) Conversely, two finite-dimensional $\mathbb{K}$-vector spaces of equal dimension are isomorphic. In particular, every finite-dimensional $\mathbb{K}$-vector space of dimension $n$ is isomorphic to $\mathbb{K}^n$.

Proof

(i) Let $u : E \to F$ be an isomorphism and $(e_1, \ldots, e_n)$ a basis of $E$. By the previous Theorem applied to $u$ bijective, $(u(e_1), \ldots, u(e_n))$ is a basis of $F$. So $F$ has a basis of cardinal $n$, hence $F$ is finite-dimensional with $\dim F = n = \dim E$.
(ii) Suppose $\dim E = \dim F = n$. Pick a basis $(e_1, \ldots, e_n)$ of $E$ and a basis $(f_1, \ldots, f_n)$ of $F$. By the « basis $\mapsto$ image of basis » Theorem applied to the family $(f_1, \ldots, f_n)$, there exists a unique linear map $u : E \to F$ with $u(e_i) = f_i$. By the previous Theorem, $u$ is bijective; hence $u$ is an isomorphism.
In particular, every $E$ of dimension $n$ is isomorphic to $\mathbb{K}^n$ (which has dimension $n$ via the canonical basis).

Skills to practice

Using isomorphism to compute a dimension

II.4 Dimension of $\mathcal{L}(E\virgule\ F)$

Proposition — Dimension of $\mathcal{L}(E\virgule\ F)$

Let $E$ and $F$ be $\mathbb{K}$-vector spaces of finite dimension. Then $\mathcal{L}(E, F)$ is finite-dimensional and $$ \textcolor{colorprop}{\dim \mathcal{L}(E, F) = \dim E \times \dim F}. $$

Proof

Set $p = \dim E$ and fix a basis $(e_1, \ldots, e_p)$ of $E$. By the « basis $\mapsto$ image of basis » Theorem, the map $$ \Phi : \mathcal{L}(E, F) \longrightarrow F^p, \qquad u \longmapsto (u(e_1), \ldots, u(e_p)) $$ is a bijection (existence + uniqueness of the linear extension of any tuple of images). It is also linear: $\Phi(\lambda u + v) = ((\lambda u + v)(e_1), \ldots) = (\lambda u(e_1) + v(e_1), \ldots) = \lambda \Phi(u) + \Phi(v)$. So $\Phi$ is an isomorphism between $\mathcal{L}(E, F)$ and $F^p$. By the previous Theorem, $\dim \mathcal{L}(E, F) = \dim F^p = p \dim F$ (dimension-of-product result from Finite-dimensional vector spaces) $= \dim E \times \dim F$.

Example

$\dim \mathcal{L}(\mathbb{R}^3, \mathbb{R}^2) = 3 \times 2 = 6$.
$\dim \mathcal{L}(\mathbb{R}_3[X], \mathbb{R}^2) = 4 \times 2 = 8$ (since $\dim \mathbb{R}_3[X] = 4$).
$\dim \mathcal{L}(E) = (\dim E)^2$ for any finite-dimensional $E$.

Skills to practice

Computing the dimension of $\mathcal{L}(E\virgule\ F)$

III Linear maps in finite dimension: rank theorem

III.1 Rank of a linear map

For a linear map of finite rank, define the rank as the dimension of the image. The rank captures how much of the target $F$ is « actually reached » by $f$. When the source $E$ is finite-dimensional, the rank is automatically bounded by $\dim E$ (one cannot create more dimensions than one starts with) and by $\dim F$ (one cannot reach more than the target offers).

Definition — Rank of a linear map

Let $f \in \mathcal{L}(E, F)$. We say that $f$ is of finite rank if $\mathrm{Im}\,f$ is a finite-dimensional subspace of $F$. In that case, the rank of $f$ is $$ \textcolor{colordef}{\mathrm{rg}(f) := \dim \mathrm{Im}\,f}. $$

Proposition — Rank bounds

Let $f \in \mathcal{L}(E, F)$ be of finite rank. Then:

(a) If $F$ is finite-dimensional, $\textcolor{colorprop}{\mathrm{rg}(f) \le \dim F}$, with equality iff $f$ is surjective.
(b) If $E$ is finite-dimensional, then $f$ is automatically of finite rank and $\textcolor{colorprop}{\mathrm{rg}(f) \le \dim E}$.

(The « equality iff $f$ injective » half of (b) follows from the rank theorem below and is stated as a Corollary after the rank theorem in this subsection.)

Proof

(a) $\mathrm{Im}\,f \subset F$ and both are finite-dimensional, so $\mathrm{rg}(f) = \dim \mathrm{Im}\,f \le \dim F$ (subspace-dimension theorem of Finite-dimensional vector spaces), with equality iff $\mathrm{Im}\,f = F$, i.e. iff $f$ is surjective.
(b) If $E$ has a basis $(e_1, \ldots, e_n)$, then $\mathrm{Im}\,f = \mathrm{Vect}(f(e_1), \ldots, f(e_n))$ is generated by $n$ vectors, so finite-dimensional with $\mathrm{rg}(f) \le n = \dim E$.

Example

Compute $\mathrm{rg}(f)$ for $f : \mathbb{R}^3 \to \mathbb{R}^3$, $(x, y, z) \mapsto (x + y,\ y + z,\ x + z)$.
$\mathrm{Im}\,f = \mathrm{Vect}\bigl(f(e_1), f(e_2), f(e_3)\bigr) = \mathrm{Vect}\bigl((1, 0, 1), (1, 1, 0), (0, 1, 1)\bigr)$. To extract a basis: $(1, 0, 1)$ and $(1, 1, 0)$ are not proportional, hence free. Is $(0, 1, 1)$ in their span? Suppose $$ (0, 1, 1) = a (1, 0, 1) + b (1, 1, 0) = (a + b,\ b,\ a). $$ Then $b = 1$ and $a = 1$, hence $a + b = 2$, contradicting the first coordinate $0$. So $(0, 1, 1) \notin \mathrm{Vect}\bigl((1, 0, 1), (1, 1, 0)\bigr)$. The three vectors are free, so $\mathrm{rg}(f) = 3$ and $f$ is surjective (and bijective).

Proposition — Rank under composition

Let $f \in \mathcal{L}(E, F)$ and $g \in \mathcal{L}(F, G)$, both of finite rank. Then $g \circ f$ is of finite rank and $$ \textcolor{colorprop}{\mathrm{rg}(g \circ f) \le \min\bigl(\mathrm{rg}(f),\ \mathrm{rg}(g)\bigr)}. $$ Moreover:

if $g_{|\mathrm{Im}\,f}$ is injective, then $\mathrm{rg}(g \circ f) = \mathrm{rg}(f)$;
if $f$ is surjective, then $\mathrm{rg}(g \circ f) = \mathrm{rg}(g)$.

Proof

Bound by $\mathrm{rg}(g)$. $\mathrm{Im}(g \circ f) = g(\mathrm{Im}\,f) \subset \mathrm{Im}\,g$, so $\mathrm{rg}(g \circ f) \le \mathrm{rg}(g)$.
Bound by $\mathrm{rg}(f)$. The restriction $g_{|\mathrm{Im}\,f} : \mathrm{Im}\,f \to G$ has $\mathrm{Im}\,(g_{|\mathrm{Im}\,f}) = g(\mathrm{Im}\,f) = \mathrm{Im}(g \circ f)$. Apply the rank-bound (b) of « Rank bounds » to $g_{|\mathrm{Im}\,f}$ (whose source is $\mathrm{Im}\,f$, of dimension $\mathrm{rg}(f)$): $\mathrm{rg}(g_{|\mathrm{Im}\,f}) \le \dim \mathrm{Im}\,f$. Since $\mathrm{rg}(g_{|\mathrm{Im}\,f}) = \dim \mathrm{Im}(g \circ f) = \mathrm{rg}(g \circ f)$ and $\dim \mathrm{Im}\,f = \mathrm{rg}(f)$, we get $\mathrm{rg}(g \circ f) \le \mathrm{rg}(f)$.

If $g_{|\mathrm{Im}\,f}$ is injective, then $g_{|\mathrm{Im}\,f} : \mathrm{Im}\,f \to \mathrm{Im}(g \circ f)$ is bijective (it is surjective onto its image by construction). So $\dim \mathrm{Im}(g \circ f) = \dim \mathrm{Im}\,f$, i.e. $\mathrm{rg}(g \circ f) = \mathrm{rg}(f)$.
If $f$ is surjective, then $\mathrm{Im}\,f = F$, so $\mathrm{Im}(g \circ f) = g(F) = \mathrm{Im}\,g$, hence $\mathrm{rg}(g \circ f) = \mathrm{rg}(g)$.

Skills to practice

Computing the rank of a linear map

III.2 Geometric form of the rank theorem

The geometric heart of the rank theorem is the following: the restriction of $f$ to any supplement of $\mathrm{Ker}\,f$ is an isomorphism onto $\mathrm{Im}\,f$. Once this is established (no finite-dimension assumption is needed), the rank--nullity formula falls out in one line in the finite-dimensional case.

Theorem — Geometric form of the rank theorem

Let $E, F$ be $\mathbb{K}$-vector spaces (not necessarily finite-dimensional) and let $f \in \mathcal{L}(E, F)$. Let $S$ be a supplement of $\mathrm{Ker}\,f$ in $E$ (i.e. $E = \mathrm{Ker}\,f \oplus S$). Then the restriction $f_{|S} : S \to \mathrm{Im}\,f$ is an isomorphism.

Proof

Injectivity of $f_{|S}$. $\mathrm{Ker}\,f_{|S} = \mathrm{Ker}\,f \cap S = \{0_E\}$ since $E = \mathrm{Ker}\,f \oplus S$ is a direct sum.
Surjectivity of $f_{|S}$ onto $\mathrm{Im}\,f$. Let $y \in \mathrm{Im}\,f$: pick $x \in E$ with $f(x) = y$. Decompose $x = k + s$ with $k \in \mathrm{Ker}\,f$ and $s \in S$. Then $y = f(x) = f(k) + f(s) = 0 + f(s) = f(s) = f_{|S}(s)$. So every $y \in \mathrm{Im}\,f$ is reached by $f_{|S}$.

Both properties together: $f_{|S} : S \to \mathrm{Im}\,f$ is bijective, hence an isomorphism (its inverse is automatically linear).

Skills to practice

Applying the geometric form of the rank theorem

III.3 Rank theorem

Theorem — Rank theorem

Let $E$ be a $\mathbb{K}$-vector space of finite dimension and let $f \in \mathcal{L}(E, F)$. Then $\mathrm{Im}\,f$ is finite-dimensional and $$ \boldsymbol{\dim E = \dim \mathrm{Ker}\,f + \mathrm{rg}(f)}. $$

Proof

$\mathrm{Ker}\,f$ is a subspace of $E$, finite-dimensional. By the supplement-existence theorem (Finite-dimensional vector spaces), $\mathrm{Ker}\,f$ admits a supplement $S$ in $E$: $E = \mathrm{Ker}\,f \oplus S$, with $\dim S = \dim E - \dim \mathrm{Ker}\,f$ (sum-of-dimensions for direct sums). By the geometric form (previous Theorem), $f_{|S} : S \to \mathrm{Im}\,f$ is an isomorphism, so $\mathrm{Im}\,f$ is finite-dimensional with $\dim \mathrm{Im}\,f = \dim S$. Combining, $$ \mathrm{rg}(f) = \dim \mathrm{Im}\,f = \dim S = \dim E - \dim \mathrm{Ker}\,f. $$

Corollary — Rank equals $\dim E$ iff $f$ injective

Let $E$ be of finite dimension and $f \in \mathcal{L}(E, F)$. Then $$ \mathrm{rg}(f) = \dim E \iff f \text{ is \textcolor{colorprop}{injective}}. $$

Proof

By the rank theorem, $\mathrm{rg}(f) = \dim E - \dim \mathrm{Ker}\,f$. So $\mathrm{rg}(f) = \dim E$ $\iff$ $\dim \mathrm{Ker}\,f = 0$ $\iff$ $\mathrm{Ker}\,f = \{0_E\}$ $\iff$ $f$ injective.

Method — Computing $\dim \mathrm{Im}\ f$ via the rank theorem

When $E$ is finite-dimensional, the rank theorem gives the shortcut $\dim \mathrm{Im}\,f = \dim E - \dim \mathrm{Ker}\,f$. So computing $\mathrm{Ker}\,f$ (solving the homogeneous linear system) is enough to read off the rank, without exhibiting a basis of $\mathrm{Im}\,f$.

Example

For $f : \mathbb{R}^3 \to \mathbb{R}^2$, $(x, y, z) \mapsto (2x + y - z,\ x - y)$ (Example above): we found $\dim \mathrm{Ker}\,f = 1$. By the rank theorem, $\mathrm{rg}(f) = 3 - 1 = 2 = \dim \mathbb{R}^2$, so $f$ is surjective --- consistent with the direct computation $\mathrm{Im}\,f = \mathbb{R}^2$.

Example

Let $u : \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $u(x, y, z) = (x + 2y - z,\ 2x - y + 8z,\ x + y + z)$.
Kernel. Solving the homogeneous system $\{x + 2y - z = 0,\ 2x - y + 8z = 0,\ x + y + z = 0\}$ (Gauss elimination): from $L_1 - L_3$, $y - 2z = 0$, so $y = 2z$; from $L_3$, $x + 2z + z = -3z$, so $x = -3z$; checking $L_2$: $2(-3z) - 2z + 8z = 0$ $\checkmark$. Hence $\mathrm{Ker}\,u = \mathrm{Vect}\bigl((-3, 2, 1)\bigr)$ and $\dim \mathrm{Ker}\,u = 1$.
Rank. By the rank theorem, $\mathrm{rg}(u) = 3 - 1 = 2$. So $u$ is neither injective nor surjective; $\mathrm{Im}\,u$ is a plane of $\mathbb{R}^3$.

Skills to practice

Applying the rank theorem
Bounding the rank under composition

III.4 Equivalence inj $\iff$ surj $\iff$ bij in equal finite dimension

When source and target have the same finite dimension, the three properties --- injective, surjective, bijective --- collapse into one. Checking the easiest of the three is enough; this dramatic short-cut is one of the most used results of the chapter.

Theorem — Injective $\iff$ surjective $\iff$ bijective in equal finite dimension

Let $E$ and $F$ be finite-dimensional $\mathbb{K}$-vector spaces with $\dim E = \dim F$, and $f \in \mathcal{L}(E, F)$. Then $$ \boldsymbol{f \text{ injective} \iff f \text{ surjective} \iff f \text{ bijective}}. $$

Proof

By the rank theorem, $\mathrm{rg}(f) = \dim E - \dim \mathrm{Ker}\,f$.

$f$ injective $\iff$ $\dim \mathrm{Ker}\,f = 0$ $\iff$ $\mathrm{rg}(f) = \dim E = \dim F$ $\iff$ $f$ surjective.
$f$ bijective $\iff$ $f$ injective and surjective $\iff$ (by the chain above) $f$ injective $\iff$ $f$ surjective.

Theorem — Left/right invertibility in finite dimension

Let $E$ be of finite dimension and $f \in \mathcal{L}(E)$. Then $$ f \in \mathrm{GL}(E) \iff f \text{ is \textcolor{colorprop}{left-invertible} in } \mathcal{L}(E) \iff f \text{ is \textcolor{colorprop}{right-invertible} in } \mathcal{L}(E). $$

Proof

If $g \circ f = \mathrm{Id}_E$ (left-invertible), then $f$ is injective: $f(x) = 0 \Rightarrow g(f(x)) = g(0) = 0 \Rightarrow x = (g \circ f)(x) = 0$. By the previous Theorem (since $\dim E = \dim E$), $f$ is bijective, hence in $\mathrm{GL}(E)$. The same $g$ is forced to equal $f^{-1}$ (right-multiply $g \circ f = \mathrm{Id}_E$ by $f^{-1}$).
If $f \circ h = \mathrm{Id}_E$ (right-invertible), then $f$ is surjective. By the previous Theorem, $f$ is bijective. Similarly $h = f^{-1}$.
Conversely, $f \in \mathrm{GL}(E)$ gives both left and right inverses (both equal $f^{-1}$).

Method — Halving the work in equal finite dimension

When $\dim E = \dim F$ (finite), check whichever of « injective » or « surjective » is easier --- bijectivity follows. For an endomorphism of $E$ (dim finite), checking injectivity alone (i.e. $\mathrm{Ker}\,f = \{0_E\}$) already proves $f \in \mathrm{GL}(E)$.

Example

The map $\varphi : \mathbb{R}^3 \to \mathbb{R}^3$, $(x, y, z) \mapsto (x + y,\ -x + y,\ z)$ is an automorphism of $\mathbb{R}^3$.
Proof. $\varphi$ is linear (each coordinate is a linear combination of $x, y, z$). Kernel: solve $\{x + y = 0,\ -x + y = 0,\ z = 0\}$, giving $x = y = z = 0$, hence $\mathrm{Ker}\,\varphi = \{0\}$. So $\varphi$ is injective; as $\dim \mathbb{R}^3 = \dim \mathbb{R}^3$, $\varphi$ is bijective.

Example

Let $\psi : \mathbb{K}_n[X] \to \mathbb{K}_n[X]$, $P \mapsto X P' + P(0)$ (with $n \ge 1$).
Well-defined and linear. If $\deg P \le n$, then $\deg P' \le n - 1$, so $\deg (X P') \le n$; $P(0)$ is a constant. Hence $X P' + P(0) \in \mathbb{K}_n[X]$, and linearity is direct.
Injective. Suppose $\psi(P) = 0$, i.e. $X P' + P(0) = 0$. Then $X P' = -P(0)$ is a constant; but $X P'$ has no constant term (it vanishes at $0$), so $P(0) = 0$ and $X P' = 0$, hence $P' = 0$, so $P$ is constant. Combined with $P(0) = 0$: $P = 0$.
By the « equal dim » theorem, $\psi$ is bijective, hence an automorphism of $\mathbb{K}_n[X]$.

Skills to practice

Using the inj $\iff$ surj $\iff$ bij equivalence

IV Endomorphisms: projectors and symmetries

IV.1 Projectors

A projector is the geometric idea of « projecting onto a subspace, in the direction of a complementary subspace ». Concretely: given $E = F \oplus G$, the projector on $F$ along $G$ keeps the $F$-part and kills the $G$-part. The algebraic characterisation $p^2 = p$ captures projectors exactly, and the image/kernel decomposition $\mathrm{Im}\,p \oplus \mathrm{Ker}\,p = E$ recovers the original splitting.

Definition — Projector

Let $E$ be a $\mathbb{K}$-vector space and $F, G$ two supplementary subspaces of $E$ (i.e. $E = F \oplus G$). The projection on $F$ along $G$ (or projector on $F$ in the direction $G$) is the map $p : E \to E$ that to $x = f + g$ ($f \in F$, $g \in G$, unique decomposition) associates $$ p(x) := f. $$ Immediate properties (by uniqueness of the decomposition):

$p$ is linear (Theorem « linear map determined by restrictions to a direct sum » applied to $p_{|F} = \mathrm{Id}_F$ and $p_{|G} = 0$);
$p^2 = p$ ($p(f) = f$ for $f \in F$, so $p(p(x)) = p(f) = f = p(x)$);
$F = \mathrm{Im}\,p = \mathrm{Ker}(p - \mathrm{Id}_E) = \{x \in E : p(x) = x\}$;
$G = \mathrm{Ker}\,p$.

Theorem — Algebraic characterisation of projectors

Let $p : E \to E$ be a map. Then $p$ is a projector if and only if $p$ is linear and $p^2 = p$. In that case, $\mathrm{Im}\,p$ and $\mathrm{Ker}\,p$ are supplementary in $E$, and $p$ is the projection on $\mathrm{Im}\,p$ along $\mathrm{Ker}\,p$. The concrete decomposition is $$ \forall x \in E : \quad x = \underbrace{p(x)}_{\in \mathrm{Im}\,p} + \underbrace{(x - p(x))}_{\in \mathrm{Ker}\,p}. $$

Proof

$(\Rightarrow)$ If $p$ is a projector for some decomposition $E = F \oplus G$, the immediate properties listed in the Definition give $p$ linear and $p^2 = p$.
$(\Leftarrow)$ Suppose $p$ linear and $p^2 = p$. For every $x \in E$, write $x = p(x) + (x - p(x))$. Then $p(x) \in \mathrm{Im}\,p$, and $p(x - p(x)) = p(x) - p^2(x) = p(x) - p(x) = 0$, so $x - p(x) \in \mathrm{Ker}\,p$. Hence $E = \mathrm{Im}\,p + \mathrm{Ker}\,p$.
Direct sum. Let $y \in \mathrm{Im}\,p \cap \mathrm{Ker}\,p$. Write $y = p(z)$ for some $z$; then $0 = p(y) = p(p(z)) = p^2(z) = p(z) = y$. So $\mathrm{Im}\,p \cap \mathrm{Ker}\,p = \{0\}$, hence $E = \mathrm{Im}\,p \oplus \mathrm{Ker}\,p$.
The decomposition above shows that $p$ keeps the $\mathrm{Im}\,p$-part of $x$, so $p$ is the projector on $\mathrm{Im}\,p$ along $\mathrm{Ker}\,p$.

Method — Showing an endomorphism is a projector

To prove $p \in \mathcal{L}(E)$ is a projector: verify $p^2 = p$ on the generic input. The algebraic characterisation Theorem then automatically gives the supplementary subspaces $F = \mathrm{Im}\,p$ and $G = \mathrm{Ker}\,p$.

Example — Projector on the plane $z = 0$ of $\mathbb{R}^3$

Consider $p : \mathbb{R}^3 \to \mathbb{R}^3$, $(x, y, z) \mapsto (x, y, 0)$.
Linearity. Each component of $p(x, y, z)$ is a linear combination of $x, y, z$.
$p^2 = p$. $p \circ p (x, y, z) = p(x, y, 0) = (x, y, 0) = p(x, y, z)$.
By the algebraic characterisation of projectors, $p$ is a projector.
Image and kernel. $\mathrm{Im}\,p = \{(x, y, 0) : x, y \in \mathbb{R}\}$ (the plane $z = 0$); $\mathrm{Ker}\,p = \{(0, 0, z) : z \in \mathbb{R}\}$ (the axis $Oz$). So $p$ is the projection on the plane $z = 0$ along the axis $Oz$.

Skills to practice

Showing $p^2 = p$ characterises a projector
Determining $\mathrm{Im}$ and $\mathrm{Ker}$ of a projector

IV.2 Symmetries

Definition — Symmetry

Let $E = F \oplus G$ and let $x = f + g$ ($f \in F$, $g \in G$) be the unique decomposition. The symmetry with respect to $F$ in the direction $G$ is the map $s : E \to E$ defined by $$ s(x) := f - g. $$ Immediate properties:

$s$ is a linear automorphism of $E$ with $s^2 = \mathrm{Id}_E$ (so $s^{-1} = s$);
$F = \mathrm{Ker}(s - \mathrm{Id}_E) = \{x \in E : s(x) = x\}$;
$G = \mathrm{Ker}(s + \mathrm{Id}_E) = \{x \in E : s(x) = -x\}$.

Theorem — Algebraic characterisation of symmetries

Let $s : E \to E$ be a map. Then $s$ is a symmetry if and only if $s$ is linear and $s^2 = \mathrm{Id}_E$. In that case, $\mathrm{Ker}(s - \mathrm{Id}_E)$ and $\mathrm{Ker}(s + \mathrm{Id}_E)$ are supplementary in $E$, and $s$ is the symmetry with respect to $\mathrm{Ker}(s - \mathrm{Id}_E)$ in the direction $\mathrm{Ker}(s + \mathrm{Id}_E)$.

Proof

$(\Rightarrow)$ The Definition above gives $s$ linear and $s^2 = \mathrm{Id}_E$.
$(\Leftarrow)$ Suppose $s$ linear and $s^2 = \mathrm{Id}_E$. Set $F := \mathrm{Ker}(s - \mathrm{Id}_E)$ and $G := \mathrm{Ker}(s + \mathrm{Id}_E)$, two subspaces of $E$. Decompose every $x \in E$ as $$ x = \underbrace{\frac{x + s(x)}{2}}_{\text{call } f} + \underbrace{\frac{x - s(x)}{2}}_{\text{call } g}. $$ Since $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ (chapter-opener convention), division by $2$ is allowed. Check $f \in F$: $s(f) = \tfrac{1}{2}(s(x) + s^2(x)) = \tfrac{1}{2}(s(x) + x) = f$, so $(s - \mathrm{Id})(f) = 0$. Check $g \in G$: $s(g) = \tfrac{1}{2}(s(x) - s^2(x)) = \tfrac{1}{2}(s(x) - x) = -g$, so $(s + \mathrm{Id})(g) = 0$. Hence $E = F + G$.
Direct sum. If $y \in F \cap G$, then $s(y) = y$ and $s(y) = -y$, so $y = -y$ hence $2y = 0$, hence $y = 0$ (division by $2$). So $F \cap G = \{0\}$, $E = F \oplus G$.
The decomposition $x = f + g$ above gives $s(x) = s(f) + s(g) = f + (-g) = f - g$, so $s$ is the symmetry with respect to $F$ along $G$.

Method — Showing an endomorphism is a symmetry

To prove $s \in \mathcal{L}(E)$ is a symmetry: verify $s^2 = \mathrm{Id}_E$ on the generic input. The supplementary subspaces are then $F = \mathrm{Ker}(s - \mathrm{Id}_E)$ (fixed vectors) and $G = \mathrm{Ker}(s + \mathrm{Id}_E)$ (vectors sent to their opposite).

Example — Axial symmetry of $\mathbb{R}^2$

Consider $s : \mathbb{R}^2 \to \mathbb{R}^2$, $(x, y) \mapsto (x, -y)$.
Linearity. Each component of $s(x, y)$ is a linear combination of $x, y$.
$s^2 = \mathrm{Id}$. $s \circ s (x, y) = s(x, -y) = (x, y)$.
By the algebraic characterisation of symmetries, $s$ is a symmetry.
Eigenspaces. $\mathrm{Ker}(s - \mathrm{Id}) = \{(x, y) : (x, -y) = (x, y)\} = \{(x, 0) : x \in \mathbb{R}\}$ (the $x$-axis); $\mathrm{Ker}(s + \mathrm{Id}) = \{(x, y) : (x, -y) = (-x, -y)\} = \{(0, y) : y \in \mathbb{R}\}$ (the $y$-axis). So $s$ is the symmetry with respect to the $x$-axis in the direction of the $y$-axis.

Skills to practice

Showing $s^2 = \mathrm{Id}$ characterises a symmetry

IV.3 Link projector $\leftrightarrow$ symmetry

A projector and a symmetry on the same decomposition $E = F \oplus G$ are linked by a one-line affine relation: $s = 2p - \mathrm{Id}_E$ goes from projector to symmetry, $p = (s + \mathrm{Id}_E)/2$ goes back. This makes one available from the other in computations and arguments.

Method — Building one from the other

Projector $\to$ symmetry: if $p$ is the projector on $F$ along $G$, then $s := 2p - \mathrm{Id}_E$ is the symmetry with respect to $F$ in the direction $G$.
Symmetry $\to$ projector: if $s$ is the symmetry with respect to $F$ along $G$, then $p := (s + \mathrm{Id}_E)/2$ is the projector on $F$ in the direction $G$.
Verification (projector $\to$ symmetry). $s^2 = (2p - \mathrm{Id})^2 = 4 p^2 - 4 p + \mathrm{Id} = 4 p - 4 p + \mathrm{Id} = \mathrm{Id}$ since $p^2 = p$. So $s$ is a symmetry by the algebraic characterisation.
Verification (symmetry $\to$ projector). Conversely, if $s^2 = \mathrm{Id}_E$ and $p = (s + \mathrm{Id}_E)/2$, then $$ p^2 = \frac{(s + \mathrm{Id})^2}{4} = \frac{s^2 + 2s + \mathrm{Id}}{4} = \frac{2 \mathrm{Id} + 2s}{4} = \frac{s + \mathrm{Id}}{2} = p. $$ So $p$ is a projector.

Example — Transposition is a symmetry

On $\mathcal{M}_n(\mathbb{R})$, the transposition $t : M \mapsto M^\top$ satisfies $t^2 = \mathrm{Id}$ (transposition is involutive). So $t$ is a symmetry.
Kernel of $t - \mathrm{Id}$. $t(M) = M$ means $M = M^\top$: $M$ is symmetric. So $\mathrm{Ker}(t - \mathrm{Id}) = \mathcal{S}_n(\mathbb{R})$ (symmetric matrices).
Kernel of $t + \mathrm{Id}$. $t(M) = -M$ means $M^\top = -M$: $M$ is antisymmetric. So $\mathrm{Ker}(t + \mathrm{Id}) = \mathcal{A}_n(\mathbb{R})$ (antisymmetric matrices).
Corollary. As $\mathcal{M}_n(\mathbb{R}) = \mathrm{Ker}(t - \mathrm{Id}) \oplus \mathrm{Ker}(t + \mathrm{Id})$ by the algebraic characterisation, we recover $$ \mathcal{M}_n(\mathbb{R}) = \mathcal{S}_n(\mathbb{R}) \oplus \mathcal{A}_n(\mathbb{R}). $$

Example — Even/odd decomposition

On $\mathbb{R}^\mathbb{R}$, the map $\sigma : f \mapsto f \circ (x \mapsto -x)$ (sending $f$ to the function $x \mapsto f(-x)$) is linear, and $\sigma^2(f) = f$. So $\sigma$ is a symmetry.
$\mathrm{Ker}(\sigma - \mathrm{Id})$ = even functions $\mathcal{P}$; $\mathrm{Ker}(\sigma + \mathrm{Id})$ = odd functions $\mathcal{I}$. So $\mathbb{R}^\mathbb{R} = \mathcal{P} \oplus \mathcal{I}$. Define $$ f_{\mathrm{even}}(x) = \frac{f(x) + f(-x)}{2}, \qquad f_{\mathrm{odd}}(x) = \frac{f(x) - f(-x)}{2}. $$ Then $f = f_{\mathrm{even}} + f_{\mathrm{odd}}$.

Skills to practice

Linking projector and symmetry

V Linear forms and hyperplanes

V.1 Linear forms and coordinate forms

A linear form is a linear map valued in the base field $\mathbb{K}$. In finite dimension, once a basis is fixed, every linear form is the same kind of object: a linear combination $a_1 x_1 + \cdots + a_n x_n$ of the coordinates. The $a_i$ are unique and equal to the values of the form on the basis vectors. The systematic study of the dual space $\mathcal{L}(E, \mathbb{K})$ and dual bases is out of program (« L'étude de la dualité est hors programme »); coordinate forms appear here only as the practical tool used to write equations of hyperplanes.

Definition — Linear form

Let $E$ be a $\mathbb{K}$-vector space. A linear form on $E$ is a linear map $\varphi : E \to \mathbb{K}$.

Proposition — Coordinate forms

Let $E$ be a $\mathbb{K}$-vector space of finite dimension with basis $\mathcal{B} = (e_1, \ldots, e_n)$.

(a) For each $i \in \llbracket 1, n \rrbracket$, the map $$ \mathrm{co}_i : E \longrightarrow \mathbb{K}, \quad x = x_1 e_1 + \cdots + x_n e_n \longmapsto x_i $$ is a linear form on $E$, called the $i$-th coordinate form in $\mathcal{B}$.
(b) Every linear form $\varphi$ on $E$ is written uniquely as a linear combination of the coordinate forms: $$ \textcolor{colorprop}{\varphi(x_1 e_1 + \cdots + x_n e_n) = a_1 x_1 + \cdots + a_n x_n} \quad \text{with} \quad a_i := \varphi(e_i). $$

Proof

(a) Linearity of $\mathrm{co}_i$: the coordinate of $\lambda x + y$ on $e_i$ equals $\lambda$ times the coordinate of $x$ plus the coordinate of $y$ (uniqueness of coordinates in a basis).
(b) Existence. For $x = \sum x_i e_i$, linearity of $\varphi$ gives $\varphi(x) = \sum x_i \varphi(e_i) = \sum a_i x_i$ with $a_i := \varphi(e_i)$.
Uniqueness. If $\varphi(x) = b_1 x_1 + \cdots + b_n x_n$ for all $x$, then evaluating at $e_j$ gives $\varphi(e_j) = b_j$. So $b_j = a_j$ for every $j$.

Out of program --- duality

The systematic study of the dual space $\mathcal{L}(E, \mathbb{K})$ and dual bases is out of program: « L'étude de la dualité est hors programme. » We use coordinate forms operationally --- only as a way to write equations of hyperplanes below.

Method — Identifying a linear form on $\mathbb{K}^n$

Every linear form on $\mathbb{K}^n$ is of the type $(x_1, \ldots, x_n) \mapsto a_1 x_1 + \cdots + a_n x_n$ for a unique tuple $(a_1, \ldots, a_n) \in \mathbb{K}^n$. Read off the coefficients directly via $a_i = \varphi(e_i)$ on the canonical basis.

Example

On $\mathbb{R}^3$, define $\varphi(x, y, z) = 2x + y - 3z$. The coefficient tuple is $(a_1, a_2, a_3) = (2, 1, -3)$, with $a_i = \varphi(e_i)$. As $\varphi$ is non-zero, its kernel $\{(x, y, z) : 2x + y - 3z = 0\}$ is a hyperplane of $\mathbb{R}^3$ (see below).

Example

On $\mathbb{R}_2[X]$, consider the three linear forms $\varphi_0 : P \mapsto P(0)$, $\varphi_1 : P \mapsto P(1)$, $\delta : P \mapsto P'(0)$. Together, they determine a polynomial: the three conditions $P(0) = 0$, $P(1) = 0$, $P'(0) = 0$ force $P = 0$ in $\mathbb{R}_2[X]$.
Proof sketch. Write $P = a + b X + c X^2$. Then $P(0) = a$, $P'(0) = b$, $P(1) = a + b + c$. The three conditions give $a = 0$, $b = 0$, $a + b + c = c = 0$. So $P = 0$.

Skills to practice

Writing coordinate forms in a given basis

V.2 Hyperplanes

Definition — Hyperplane

Let $E$ be a $\mathbb{K}$-vector space. A hyperplane of $E$ is the kernel of a non-zero linear form on $E$: $$ H = \mathrm{Ker}\,\varphi \quad \text{with} \quad \varphi \in \mathcal{L}(E, \mathbb{K}) \setminus \{0\}. $$

Theorem — Geometric characterisation of hyperplanes

Let $H$ be a subspace of $E$. The following are equivalent:

(i) $H$ is a hyperplane of $E$;
(ii) $H$ admits a supplement that is a line, i.e. there exists $v \in E$ with $E = H \oplus \mathrm{Vect}(v)$.

Finite-dimensional case. If $\dim E = n \ge 1$, hyperplanes of $E$ are exactly the subspaces of $E$ of dimension $n - 1$.

Proof

(i) $\Rightarrow$ (ii). Suppose $H = \mathrm{Ker}\,\varphi$ with $\varphi \ne 0$. Pick $v \in E$ with $\varphi(v) \ne 0$. We show $E = H \oplus \mathrm{Vect}(v)$.
- Sum is $E$. For $x \in E$, write $x = \bigl(x - \tfrac{\varphi(x)}{\varphi(v)} v\bigr) + \tfrac{\varphi(x)}{\varphi(v)} v$. Applying $\varphi$ to the first piece, by linearity: $\varphi\bigl(x - \tfrac{\varphi(x)}{\varphi(v)} v\bigr) = \varphi(x) - \tfrac{\varphi(x)}{\varphi(v)} \varphi(v) = \varphi(x) - \varphi(x) = 0$, so the first piece is in $H$; the second is clearly in $\mathrm{Vect}(v)$.
- Direct sum. If $y \in H \cap \mathrm{Vect}(v)$, write $y = \alpha v$. Then $0 = \varphi(y) = \alpha \varphi(v)$, hence $\alpha = 0$ (since $\varphi(v) \ne 0$), hence $y = 0$.
(ii) $\Rightarrow$ (i). Suppose $E = H \oplus \mathrm{Vect}(v)$ for some $v \ne 0$. Define $\varphi : E \to \mathbb{K}$ on the direct sum by $\varphi_{|H} = 0$ and $\varphi(v) = 1$; the « restrictions to a direct sum » Theorem above gives a unique linear $\varphi$. Then $\varphi$ is non-zero ($\varphi(v) = 1$), and $\mathrm{Ker}\,\varphi = H$ since any $x = h + \alpha v$ satisfies $\varphi(x) = \alpha$, which is $0$ iff $\alpha = 0$ iff $x \in H$.

Finite-dimensional case. If $E = H \oplus \mathrm{Vect}(v)$ with $\dim E = n$, then $\dim \mathrm{Vect}(v) = 1$, so $\dim H = n - 1$. Conversely, suppose $\dim H = n - 1$. Since $\dim H < n$, $H \ne E$, so a $v \notin H$ exists. Then $H \subsetneq H + \mathrm{Vect}(v) \subset E$, so $\dim(H + \mathrm{Vect}(v)) = n$, hence $E = H + \mathrm{Vect}(v)$; and $H \cap \mathrm{Vect}(v) = \{0\}$ since $v \notin H$. So $E = H \oplus \mathrm{Vect}(v)$.

Method — Recognising a hyperplane

In finite dimension $n$, a hyperplane is a subspace of codimension $1$, i.e. of dimension $n - 1$. Equivalently: the set of solutions of a single non-trivial linear equation in the coordinates.

Example

In $\mathbb{R}^4$, the set $H = \{(x, y, z, t) : 2 x + y - z + t = 0\}$ is a hyperplane of dimension $3$. It is the kernel of the non-zero linear form $\varphi(x, y, z, t) = 2 x + y - z + t$.

Example

In $\mathbb{R}_3[X]$ (dimension $4$), the set $H = \{P \in \mathbb{R}_3[X] : P(0) = P(1)\}$ is a hyperplane: it is the kernel of the non-zero linear form $\varphi : P \mapsto P(1) - P(0)$, so $\dim H = 3$.

Skills to practice

Identifying a hyperplane

V.3 Equations of a hyperplane: comparison

Proposition — Two equations of the same hyperplane are proportional

Let $H$ be a hyperplane of $E$ and let $\varphi, \psi$ be two non-zero linear forms of $E$ whose kernel is $H$. Then there exists $\lambda \in \mathbb{K}^*$ such that $$ \textcolor{colorprop}{\psi = \lambda \varphi}. $$ In other words, two equations of the same hyperplane are scalar multiples of each other.

Proof

Pick $v \notin H$ (exists since $H \ne E$). Then $\varphi(v) \ne 0$ and $\psi(v) \ne 0$. Set $\lambda := \psi(v) / \varphi(v) \in \mathbb{K}^*$ and consider $\psi - \lambda \varphi$:

on $H$: $\psi - \lambda \varphi$ vanishes (both $\psi$ and $\varphi$ do);
at $v$: $(\psi - \lambda \varphi)(v) = \psi(v) - \lambda \varphi(v) = \psi(v) - \psi(v) = 0$.

By the previous Theorem $E = H \oplus \mathrm{Vect}(v)$, so $\psi - \lambda \varphi$ vanishes on a generating family of $E$, hence on $E$. So $\psi = \lambda \varphi$.

Skills to practice

Comparing two equations of a hyperplane

V.4 Intersection of hyperplanes

Theorem — Intersection of $p$ hyperplanes

Let $E$ be a $\mathbb{K}$-vector space of finite dimension $n \ge 1$, and let $p \in \llbracket 1, n \rrbracket$.

(i) The intersection of $p$ hyperplanes of $E$ is a subspace of $E$ of dimension at least $n - p$.
(ii) Conversely, every subspace of $E$ of dimension $n - p$ is the intersection of $p$ hyperplanes of $E$.

Proof

(i) Let $H_1, \ldots, H_p$ be $p$ hyperplanes of $E$, with equations $\varphi_1, \ldots, \varphi_p$ (non-zero linear forms). Consider the linear map $$ \Phi : E \longrightarrow \mathbb{K}^p, \quad x \longmapsto (\varphi_1(x), \ldots, \varphi_p(x)). $$ Then $\mathrm{Ker}\,\Phi = \bigcap_{i=1}^p \mathrm{Ker}\,\varphi_i = \bigcap H_i$. By the rank theorem, $\dim \mathrm{Ker}\,\Phi = n - \mathrm{rg}(\Phi) \ge n - p$ since $\mathrm{rg}(\Phi) \le \dim \mathbb{K}^p = p$.
(ii) Let $G$ be a subspace of $E$ of dimension $n - p$. Choose a basis $(e_1, \ldots, e_{n-p})$ of $G$ and complete it (theorem of the incomplete basis, prerequisite Finite-dimensional vector spaces) into a basis $(e_1, \ldots, e_n)$ of $E$. In this basis, a vector $x = \sum x_i e_i$ lies in $G$ if and only if $x_{n-p+1} = \cdots = x_n = 0$. The $p$ coordinate forms $\mathrm{co}_{n-p+1}, \ldots, \mathrm{co}_n$ from Proposition « coordinate forms » are non-zero linear forms, so their kernels are $p$ hyperplanes of $E$. Their intersection is exactly $G$.

Method — Describing a subspace by linear equations

To describe a subspace $G$ of $E$ (dim $n$) of dim $n - p$: write a system of $p$ linearly independent linear equations in the coordinates. Each equation defines a hyperplane; $G$ is their intersection. Conversely, the dimension of the solution space of a $p$-equation linear system in $n$ unknowns is at least $n - p$, with equality when the $p$ equations are linearly independent.

Example

In $\mathbb{R}^3$, the line $D = \{(x, y, z) : x = y \text{ and } z = 0\}$ is the intersection of two hyperplanes (planes): $H_1 = \{x - y = 0\}$ and $H_2 = \{z = 0\}$. By the Theorem, $\dim D \ge 3 - 2 = 1$. Conversely $D$ is parametrised by $(t, t, 0)$, so $\dim D = 1$ exactly.

Skills to practice

Intersection of hyperplanes

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