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Vector spaces

⌚ ~123 min ▢ 15 blocks ✓ 55 exercises ➣ Prerequisites : Vector Geometry of Lines and Planes

A vector space is an algebraic structure built on one operation: the linear combination. Many familiar sets already carry this structure: the plane $\mathbb{R}^2$ and space $\mathbb{R}^3$ of geometric vectors, the column space $\mathbb{K}^n$, the matrix space $M_{n,p}(\mathbb{K})$, the polynomial space $\mathbb{K}[X]$, the function space $\mathbb{K}^\Omega$, the sequence space $\mathbb{K}^\mathbb{N}$. The trick of this chapter is to isolate the common structure --- the axioms of a vector space --- so that every theorem proved about an abstract vector space is automatically true about each of these concrete instances.
Three pedagogical questions guide the chapter. Given a family of vectors $(v_1, \ldots, v_p)$: does every vector of the space write as a linear combination of the family (existence --- the family is generating) ? Is that linear combination unique (uniqueness --- the family is free) ? Both at once (existence and uniqueness --- the family is a basis) ? The motivation example below makes this concrete in $\mathbb{R}^2$.

Motivation --- a system of two equations

The sum of a son's height and his father's height is $2.5$ m; the father's height minus the son's height is $0.5$ m. Letting $x$ denote the son's height and $y$ the father's, the system reads $$ \begin{cases} x + y = 2.5 \\ -x + y = 0.5 \end{cases} \quad\text{i.e.}\quad x \begin{pmatrix} 1 \\ -1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 0.5 \end{pmatrix}. $$ The question becomes: is the right-hand side a linear combination of the two column vectors --- and if so, with which coefficients? Three structural questions follow:

Existence. Is the family $\bigl((1, -1), (1, 1)\bigr)$ generating in $\mathbb{R}^2$: can every $\mathbb{R}^2$-vector be written as a linear combination of these two?
Uniqueness. Is the family free: when an $\mathbb{R}^2$-vector is written as a linear combination of these two, are the coefficients unique?
Both at once. Is the family a basis, i.e.\ both generating and free: does every $\mathbb{R}^2$-vector admit a unique decomposition?

The vocabulary « generating », « free », « basis » belongs not just to $\mathbb{R}^2$ but to any structure in which one can form linear combinations. That structure is the vector space, the subject of this chapter.

Conventions

Throughout this chapter, $\mathbb{K}$ denotes either $\mathbb{R}$ or $\mathbb{C}$. The neutral element of an additive group is written $0$; when ambiguous, $0_E$ for the zero vector of $E$, $0_\mathbb{K}$ for the zero scalar. Vectors are written without arrows: $x \in E$, not $\overrightarrow{x}$ (in linear algebra the arrow convention is dropped). Three figures introduce the four basic operations on a geometric vector: representation, homothety (scaling by a scalar), addition, linear combination.

I Vector spaces

I.1 Definition and rules of calculation

The definition unifies under one set of axioms the addition of plane vectors, the addition of columns of $\mathbb{K}^n$, the addition of matrices, the addition of polynomials, and the addition of functions or sequences --- together with the corresponding multiplication by a scalar of $\mathbb{K}$. The four axioms on the scalar multiplication are the ones one would write down after observing how scalar multiplication interacts with addition in any of those reference examples.

Definition — Vector space

A $\mathbb{K}$-vector space is a triplet $(E, +, \cdot)$ where:

$E$ is a non-empty set,
$+$ is a binary operation on $E$ such that $(E, +)$ is a commutative group, with neutral element written $0_E$,
$\cdot$ is an external operation $\mathbb{K} \times E \to E$ (called scalar multiplication) such that, for all $\lambda, \mu \in \mathbb{K}$ and all $x, y \in E$:
1. $1 \cdot x = x$,
2. $\lambda \cdot (x + y) = \lambda \cdot x + \lambda \cdot y$,
3. $(\lambda + \mu) \cdot x = \lambda \cdot x + \mu \cdot x$,
4. $\lambda \cdot (\mu \cdot x) = (\lambda \mu) \cdot x$.

Elements of $E$ are called vectors; elements of $\mathbb{K}$ are called scalars; the field $\mathbb{K}$ is the base field of $E$. When the operation $\cdot$ is unambiguous, $\lambda \cdot x$ is shortened to $\lambda x$.

Example

Three reference vector spaces: $\mathbb{R}^2$ (plane geometric vectors), $\mathbb{R}^3$ (space geometric vectors), $(\mathbb{K}, +, \times)$ viewed as a $\mathbb{K}$-vector space (the field is itself a vector space over itself, with scalar multiplication being the field multiplication).

Proposition — Rules of calculation

Let $E$ be a $\mathbb{K}$-vector space. For all $\lambda \in \mathbb{K}$ and all $x \in E$:

$0_\mathbb{K} \cdot x = 0_E$ and $\lambda \cdot 0_E = 0_E$;
$\lambda \cdot x = 0_E \iff \lambda = 0_\mathbb{K}$ or $x = 0_E$;
$(-1) \cdot x = -x$ (the opposite of $x$ in $(E, +)$).

Proof

We prove the three points in order.

Point 1. Compute $$ \begin{aligned} 0_\mathbb{K} \cdot x & = (0_\mathbb{K} + 0_\mathbb{K}) \cdot x && \text{(neutral element of $+$ in $\mathbb{K}$)} \\ & = 0_\mathbb{K} \cdot x + 0_\mathbb{K} \cdot x && \text{(distributivity, axiom 3).} \end{aligned} $$ Simplifying $0_\mathbb{K} \cdot x$ on both sides in the group $(E, +)$ yields $0_\mathbb{K} \cdot x = 0_E$. The proof of $\lambda \cdot 0_E = 0_E$ is symmetric: $\lambda \cdot 0_E = \lambda \cdot (0_E + 0_E) = \lambda \cdot 0_E + \lambda \cdot 0_E$, simplify.
Point 2. If $\lambda = 0_\mathbb{K}$ or $x = 0_E$ then $\lambda \cdot x = 0_E$ by point 1. Conversely, suppose $\lambda \cdot x = 0_E$ with $\lambda \ne 0_\mathbb{K}$. Multiplying by $\lambda^{-1}$ (which exists in $\mathbb{K}$ since $\mathbb{K}$ is a field): $$ x = 1 \cdot x = (\lambda^{-1} \lambda) \cdot x = \lambda^{-1} \cdot (\lambda \cdot x) = \lambda^{-1} \cdot 0_E = 0_E. $$
Point 3. Compute $$ x + (-1) \cdot x = 1 \cdot x + (-1) \cdot x = (1 + (-1)) \cdot x = 0_\mathbb{K} \cdot x = 0_E, $$ so $(-1) \cdot x$ is the opposite of $x$ in $(E, +)$, i.e.\ $(-1) \cdot x = -x$.

Skills to practice

Recognizing a vector space

I.2 Reference vector spaces

We single out five reference vector spaces that the student must recognize as such on sight: the product $E_1 \times \dots \times E_n$ of vector spaces, the space $\mathbb{K}^n$ (column space), the matrix space $M_{n,p}(\mathbb{K})$, the polynomial space $\mathbb{K}[X]$, and the function space $\mathbb{K}^\Omega$ (with the special case $\mathbb{K}^\mathbb{N}$ of sequences). Two general constructions deliver all of them: the product construction and the function construction. Both verify the axioms once and for all.

Proposition — Product vector space

Let $E_1, \ldots, E_n$ be $\mathbb{K}$-vector spaces. The Cartesian product $E_1 \times \dots \times E_n$, equipped with the componentwise operations $$ (x_1, \ldots, x_n) + (y_1, \ldots, y_n) := (x_1 + y_1, \ldots, x_n + y_n), \quad \lambda \cdot (x_1, \ldots, x_n) := (\lambda \cdot x_1, \ldots, \lambda \cdot x_n), $$ is a $\mathbb{K}$-vector space, with zero element $0_{E_1 \times \dots \times E_n} = (0_{E_1}, \ldots, 0_{E_n})$.

Proof

The group structure on $E_1 \times \dots \times E_n$ is the product of group structures, which is well known. We verify two of the four axioms on the scalar multiplication; the remaining two are analogous. For $(x_1, \ldots, x_n), (y_1, \ldots, y_n) \in E_1 \times \dots \times E_n$ and $\lambda, \mu \in \mathbb{K}$: $$ \begin{aligned} 1 \cdot (x_1, \ldots, x_n) & = (1 \cdot x_1, \ldots, 1 \cdot x_n) = (x_1, \ldots, x_n), \\ \lambda \cdot \bigl((x_1, \ldots, x_n) + (y_1, \ldots, y_n)\bigr) & = \lambda \cdot (x_1 + y_1, \ldots, x_n + y_n) = (\lambda(x_1 + y_1), \ldots, \lambda(x_n + y_n)) \\ & = (\lambda x_1 + \lambda y_1, \ldots, \lambda x_n + \lambda y_n) = \lambda \cdot (x_1, \ldots, x_n) + \lambda \cdot (y_1, \ldots, y_n). \end{aligned} $$ The remaining two axioms ($(\lambda + \mu) \cdot x = \lambda x + \mu x$ and $\lambda(\mu x) = (\lambda \mu) x$) follow the same componentwise pattern.

Proposition — Vector space of functions

Let $\Omega$ be a non-empty set and $E$ a $\mathbb{K}$-vector space. The set $E^\Omega$ of functions $\Omega \to E$, equipped with the pointwise operations $$ (f + g)(t) := f(t) + g(t), \quad (\lambda \cdot f)(t) := \lambda \cdot f(t) \quad (t \in \Omega), $$ is a $\mathbb{K}$-vector space. Its zero element is the constant zero function $t \mapsto 0_E$.

Proof

Each axiom is verified by checking it pointwise on $\Omega$: for every $t \in \Omega$, the value of both sides in $E$ coincides because $E$ is a vector space. For instance, $(1 \cdot f)(t) = 1 \cdot f(t) = f(t)$ for every $t$, so $1 \cdot f = f$. Similarly, for $\lambda \in \mathbb{K}$ and $f, g \in E^\Omega$ and any $t \in \Omega$: $$ (\lambda \cdot (f + g))(t) = \lambda \cdot (f + g)(t) = \lambda \cdot (f(t) + g(t)) = \lambda f(t) + \lambda g(t) = (\lambda f + \lambda g)(t), $$ hence $\lambda \cdot (f + g) = \lambda f + \lambda g$. The two remaining axioms are analogous.

Example

The space $\mathbb{K}^n$ of $n$-tuples, $\mathbb{K}^n := \mathbb{K} \times \dots \times \mathbb{K}$ ($n$ factors), is a $\mathbb{K}$-vector space by the product construction. Addition is componentwise: $(a_1, \ldots, a_n) + (b_1, \ldots, b_n) = (a_1 + b_1, \ldots, a_n + b_n)$. Scalar multiplication is componentwise: $\lambda (a_1, \ldots, a_n) = (\lambda a_1, \ldots, \lambda a_n)$. For instance, in $\mathbb{R}^3$: $(1, 4, -3) + 2 \cdot (0, 2, 5) = (1, 8, 7)$.

Example

The matrix space $M_{n,p}(\mathbb{K})$ is a $\mathbb{K}$-vector space for its usual addition and scalar multiplication, as introduced in Matrix calculus. Indeed $M_{n,p}(\mathbb{K})$ is isomorphic to $\mathbb{K}^{np}$ (read entries row by row), so the structure is inherited from the product construction.

Example

The polynomial space $\mathbb{K}[X]$ is a $\mathbb{K}$-vector space for its usual addition and scalar multiplication (multiplying every coefficient of a polynomial by the same scalar). The zero element is the zero polynomial.

Example

The function space $\mathbb{K}^I$ (functions from an interval $I \subset \mathbb{R}$ to $\mathbb{K}$) is a $\mathbb{K}$-vector space by the function construction. The same applies to $\mathbb{K}^\mathbb{N}$ (sequences with values in $\mathbb{K}$). Addition is pointwise: $(f + g)(t) = f(t) + g(t)$; scalar multiplication is pointwise: $(\lambda f)(t) = \lambda f(t)$.

Method

To show that an ensemble $F$ is a vector space, do not verify the four scalar-multiplication axioms from scratch. Instead, find a known $\mathbb{K}$-vector space $E$ containing $F$ and prove that $F$ is a subspace of $E$ (criterion to come). Only three conditions to check then, not eight.

Skills to practice

Identifying reference spaces

I.3 Linear combinations

The central computational operation in a vector space combines $+$ and $\cdot$ into a single object: the linear combination. We always combine a finite family of vectors: a linear combination of $x_1, \ldots, x_n$ is the vector $\lambda_1 x_1 + \dots + \lambda_n x_n$ obtained by scaling each $x_k$ by a scalar $\lambda_k$ and adding the results.

Definition — Linear combination of a finite family

Let $E$ be a $\mathbb{K}$-vector space and $x_1, \ldots, x_n \in E$. A linear combination of $x_1, \ldots, x_n$ is a vector of the form $$ \sum_{k=1}^n \lambda_k x_k = \lambda_1 x_1 + \lambda_2 x_2 + \dots + \lambda_n x_n, $$ for some scalars $\lambda_1, \ldots, \lambda_n \in \mathbb{K}$. The $\lambda_k$ are called the coefficients of the linear combination.

Geometric picture of $\tfrac{1}{2} u - v$ as a linear combination of $u$ and $v$ in $\mathbb{R}^2$:

Example

Show that $(2, 7) \in \mathbb{R}^2$ is a linear combination of $u = (5, -2)$ and $v = (1, -3)$. Find explicit coefficients.

Answer

Look for $\lambda, \mu \in \mathbb{R}$ with $\lambda (5, -2) + \mu (1, -3) = (2, 7)$, i.e.\ the system $$ \begin{cases} 5 \lambda + \mu = 2 \\ -2 \lambda - 3 \mu = 7 \end{cases} $$ By Gauss ($L_2 \leftarrow 5 L_2 + 2 L_1$): $-13 \mu = 39$, so $\mu = -3$ and $\lambda = (2 - \mu)/5 = 1$. Hence $(2, 7) = 1 \cdot (5, -2) + (-3) \cdot (1, -3)$.

Example

The matrix $\begin{pmatrix} -1 & 2 \\ 2 & 0 \end{pmatrix}$ is not a linear combination of $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$, $\begin{pmatrix} -2 & 1 \\ 3 & 2 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$ in $M_2(\mathbb{R})$.

Answer

Look for $x, y, z \in \mathbb{R}$ such that $x A_1 + y A_2 + z A_3 = M$, where $A_1, A_2, A_3, M$ denote the four matrices in order. Reading entry by entry yields the system $$ \begin{cases} x - 2 y + z = -1 \\ x + y = 2 \\ 3 y - z = 2 \\ 2 y + z = 0. \end{cases} $$ Gauss elimination: apply $L_2 \leftarrow L_2 - L_1$, giving $L_2 = (0, 3, -1, 3)$. Since $L_3 = (0, 3, -1, 2)$, subtracting yields $L_3 \leftarrow L_3 - L_2 = (0, 0, 0, -1)$, i.e.\ $0 = -1$, impossible. The system has no solution: $M$ is not a linear combination of $A_1, A_2, A_3$.

Identification trap

Beware: in general, $\sum_k \lambda_k x_k = \sum_k \mu_k x_k$ does not imply $\lambda_k = \mu_k$ for all $k$. For instance in $\mathbb{R}^2$: $$ (1, 1) + 2 (0, 1) + 2 (1, 0) = (3, 3) = 2 (1, 1) + (0, 1) + (1, 0), $$ yet the coefficient triplets $(1, 2, 2)$ and $(2, 1, 1)$ are different. The right-to-identify property of coefficients will be exactly the « free family » property below.

Method

To check whether $v \in E$ is a linear combination of $x_1, \ldots, x_n$: write the equation $\lambda_1 x_1 + \dots + \lambda_n x_n = v$, identify component by component in $\mathbb{K}^n$, coefficient by coefficient in $\mathbb{K}[X]$, or entry by entry in $M_{n,p}(\mathbb{K})$, reach a linear system in the unknowns $\lambda_1, \ldots, \lambda_n$, solve it by Gauss. If the system has a solution, $v$ is a linear combination; if not, $v$ is not.

Skills to practice

Computing linear combinations

II Subspaces

II.1 Definition and characterization

A subspace of a vector space $E$ is a subset $F \subset E$ which is itself a vector space for the operations inherited from $E$. The point is twofold:

subspaces are everywhere (the solution set of a linear equation, the kernel of a linear map, an axis, a plane, the set of continuous functions, the set of bounded sequences);
verifying that $F$ is a subspace is much easier than reverifying all the axioms of a vector space --- only three conditions remain.

Definition — Subspace of a vector space

Let $E$ be a $\mathbb{K}$-vector space. A subspace of $E$ is a subset $F \subset E$ such that the restrictions to $F$ of the two operations $+$ and $\cdot$ of $E$ make $(F, +, \cdot)$ a $\mathbb{K}$-vector space.

Proposition — Characterization of a subspace

Let $E$ be a $\mathbb{K}$-vector space and $F \subset E$. The following are equivalent:

$F$ is a subspace of $E$.
$0_E \in F$ and for all $\lambda \in \mathbb{K}$ and all $x, y \in F$, $\lambda x + y \in F$ (we say $F$ is stable under linear combination).

Proof

$(1 \Rightarrow 2)$. Suppose $F$ is a subspace. Then $F$ contains its own zero, which must equal $0_E$ since $0_E + 0_F = 0_F$ in $E$ and uniqueness of the neutral element gives $0_F = 0_E$. For stability: given $\lambda \in \mathbb{K}$ and $x, y \in F$, both $\lambda x$ and $\lambda x + y$ belong to $F$ since the laws on $F$ are the restrictions of those on $E$.
$(2 \Rightarrow 1)$. Suppose $0_E \in F$ and $F$ stable under $(\lambda, x, y) \mapsto \lambda x + y$. With $\lambda = 1$ we get $x + y \in F$ for all $x, y \in F$, so $+$ restricts to $F$; with $y = 0_E$ we get $\lambda x \in F$, so $\cdot$ restricts to $F$. Associativity and commutativity of $+$, together with the four scalar-multiplication axioms, are inherited from $E$. For the group structure on $F$: the opposite of $x \in F$ is $(-1) \cdot x \in F$ (by Proposition « Rules of calculation », point 3). Hence $(F, +, \cdot)$ is a $\mathbb{K}$-vector space.

The summary slogan: « qui peut le plus peut le moins » --- stability under $\lambda x + y$ packages stability under $+$ ($\lambda = 1$), stability under $\cdot$ ($y = 0_E$), and the existence of opposites ($\lambda = -1, y = 0_E$) into a single condition.

A line through the origin (left) is a subspace; a line not through the origin (right) is not a subspace because it fails to contain $0$.

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Example

Show that $F = \{(x, y) \in \mathbb{R}^2 : x + 2 y = 0\}$ is a subspace of $\mathbb{R}^2$.

Answer

Apply the characterization.

$(0, 0) \in F$ since $0 + 2 \cdot 0 = 0$.
Take $\lambda \in \mathbb{R}$ and $u_1 = (x_1, y_1), u_2 = (x_2, y_2) \in F$ (so $x_1 + 2 y_1 = 0$ and $x_2 + 2 y_2 = 0$). Then $\lambda u_1 + u_2 = (\lambda x_1 + x_2, \lambda y_1 + y_2)$, and $$ (\lambda x_1 + x_2) + 2 (\lambda y_1 + y_2) = \lambda \underbrace{(x_1 + 2 y_1)}_{= 0} + \underbrace{(x_2 + 2 y_2)}_{= 0} = 0, $$ so $\lambda u_1 + u_2 \in F$.

Hence $F$ is a subspace of $\mathbb{R}^2$ (geometrically: a line through the origin).

Example

The set $F_1 = \{(x, y) \in \mathbb{R}^2 : x + y = 2\}$ is not a subspace of $\mathbb{R}^2$, because the zero vector $(0, 0)$ does not satisfy $0 + 0 = 2$. (Geometrically: an affine line not passing through the origin.)

Example

The set $\mathcal{C}(\mathbb{R}, \mathbb{R})$ of continuous functions $\mathbb{R} \to \mathbb{R}$ is a subspace of $\mathbb{R}^\mathbb{R}$. The set of convergent real sequences is a subspace of $\mathbb{R}^\mathbb{N}$.

Answer

The zero function is continuous; the sum of two continuous functions is continuous; a scalar multiple of a continuous function is continuous. Similarly, the zero sequence is convergent (towards $0$); the sum of two convergent sequences is convergent; a scalar multiple of a convergent sequence is convergent. Hence both sets are subspaces.

Example

The set $\mathcal{T}_n^+(\mathbb{K})$ of upper-triangular matrices of size $n$ is a subspace of $M_n(\mathbb{K})$: the zero matrix is upper-triangular, and the sum / scalar multiple of upper-triangular matrices is upper-triangular (since the entries below the diagonal stay zero).

Method

To show $F \subset E$ is a subspace: verify the three conditions of the characterization, in this order:

$0_E \in F$ (or equivalently, $F$ is non-empty and contains the zero of $E$);
for all $x, y \in F$, $x + y \in F$ (stability under $+$);
for all $\lambda \in \mathbb{K}$ and all $x \in F$, $\lambda x \in F$ (stability under $\cdot$).

The packaged form $0_E \in F$ and $\lambda x + y \in F$ is equivalent and often more compact in practice.

Skills to practice

Identifying subspaces

II.2 Reference subspaces

We single out a short list of « reference » subspaces to be recognized on sight: the trivial subspaces $\{0_E\}$ and $E$; lines through the origin in $\mathbb{R}^2$ and $\mathbb{R}^3$; planes through the origin in $\mathbb{R}^3$; $\mathbb{K}_n[X]$ as a subspace of $\mathbb{K}[X]$; the solution set of a linear homogeneous system as a subspace of $\mathbb{K}^p$.

Example

For any $\mathbb{K}$-vector space $E$, the singleton $\{0_E\}$ and the whole space $E$ are subspaces of $E$. They are called the trivial subspaces.

Example

A line through the origin of $\mathbb{R}^2$ is a subspace of the form $\{\lambda u : \lambda \in \mathbb{R}\}$ for some non-zero vector $u \in \mathbb{R}^2$ (the « direction » of the line). A plane through the origin of $\mathbb{R}^3$ is a subspace of the form $\{\lambda u + \mu v : \lambda, \mu \in \mathbb{R}\}$ for two non-collinear vectors $u, v \in \mathbb{R}^3$. Both are subspaces: they contain $0$ and are stable under linear combination.

Geometric pictures of a line through the origin in $\mathbb{R}^2$ (left) and a plane through the origin in $\mathbb{R}^3$ (right):

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Example

For every $n \in \mathbb{N}$, the set $\mathbb{K}_n[X]$ of polynomials of degree at most $n$ is a subspace of $\mathbb{K}[X]$: the zero polynomial is in $\mathbb{K}_n[X]$, and a linear combination of polynomials of degree at most $n$ has degree at most $n$.

Proposition — Solution set of a homogeneous linear system

Let $A \in M_{n,p}(\mathbb{K})$. The solution set $$ S := \{X \in \mathbb{K}^p : A X = 0_{n,1}\} $$ of the homogeneous linear system $A X = 0$ (with unknown $X \in \mathbb{K}^p$) is a subspace of $\mathbb{K}^p$.

Proof

We check stability under linear combination. First, $0_{p,1} \in S$ since $A \cdot 0_{p,1} = 0_{n,1}$. Next, for $\lambda \in \mathbb{K}$ and $X, X' \in S$ (so $A X = 0$ and $A X' = 0$), linearity of the matrix product $X \mapsto A X$ (proved in Matrix calculus) gives $$ A (\lambda X + X') = \lambda \cdot A X + A X' = \lambda \cdot 0 + 0 = 0, $$ so $\lambda X + X' \in S$. By the characterization, $S$ is a subspace of $\mathbb{K}^p$.

Skills to practice

Recognizing function subspaces

II.3 Intersection of subspaces

Subspaces behave well under intersection but not under union. The intersection of any family of subspaces is again a subspace; this is the key fact that will let us define « the smallest subspace containing a set » as an intersection. The union, by contrast, is almost never a subspace, because the sum of one vector from each piece may fall outside both.

Proposition — Intersection of subspaces

Let $E$ be a $\mathbb{K}$-vector space and $(F_i)_{i \in I}$ a non-empty family of subspaces of $E$. Then the intersection $$ \bigcap_{i \in I} F_i $$ is a subspace of $E$. (For $I = \emptyset$, we use the convention $\bigcap_{i \in \emptyset} F_i = E$, which is also a subspace.)

Proof

Set $F = \bigcap_{i \in I} F_i$. First, $0_E \in F_i$ for each $i$ (since each $F_i$ is a subspace), so $0_E \in F$. Next, for $\lambda \in \mathbb{K}$ and $x, y \in F$, we have $x, y \in F_i$ for each $i$, so $\lambda x + y \in F_i$ (each $F_i$ is stable under linear combination). Hence $\lambda x + y \in F$. By the characterization, $F$ is a subspace of $E$.

Example

The intersection of two distinct planes through the origin of $\mathbb{R}^3$ is a line through the origin of $\mathbb{R}^3$. For instance, $$ \{(x, y, z) \in \mathbb{R}^3 : x + 3 y + z = 0\} \cap \{(x, y, z) \in \mathbb{R}^3 : z = x\} $$ is a subspace of $\mathbb{R}^3$ (by the proposition) which equals $\{(x, y, x) \in \mathbb{R}^3 : 2 x + 3 y = 0\} = \mathbb{R} \cdot (3, -2, 3)$ --- a line through the origin.

Warning --- union is not a subspace

The union of two subspaces is almost never a subspace. For instance, in $\mathbb{R}^2$, take $F = \mathbb{R} \cdot (1, 0)$ (the $x$-axis) and $G = \mathbb{R} \cdot (0, 1)$ (the $y$-axis). Then $u = (1, 0) \in F \subset F \cup G$ and $v = (0, 1) \in G \subset F \cup G$, but $u + v = (1, 1) \notin F$ and $u + v \notin G$, so $u + v \notin F \cup G$. The union fails to be stable under addition.

Skills to practice

Computing intersections

II.4 Subspace spanned by a set

Given a part $X$ of a vector space $E$, the « smallest » subspace of $E$ containing $X$ exists and admits two equivalent descriptions: as an intersection (intersection of all subspaces of $E$ containing $X$) or as a set of linear combinations (all the linear combinations of elements of $X$).

Definition — Subspace spanned by a set

Let $E$ be a $\mathbb{K}$-vector space and $X \subset E$. The subspace spanned by $X$, written $\mathrm{Vect}(X)$, is the intersection of all subspaces of $E$ containing $X$: $$ \mathrm{Vect}(X) := \bigcap_{\substack{F \text{ subspace} \\ X \subset F}} F. $$ Equivalently, $\mathrm{Vect}(X)$ is the set of all linear combinations of elements of $X$: writing $X = \{x_i : i \in I\}$, $$ \mathrm{Vect}(X) = \left\{ \sum_{i \in I} \lambda_i x_i \ : \ (\lambda_i)_{i \in I} \text{ almost-zero family of } \mathbb{K} \right\}. $$ We also write $\mathrm{Vect}\bigl((x_i)_{i \in I}\bigr)$ for the same object. By convention, $\mathrm{Vect}(\emptyset) = \{0_E\}$.

Geometric pictures: $\mathrm{Vect}(u)$ is the line directed by $u$ (left); $\mathrm{Vect}(u, v)$ is the plane spanned by $u$ and $v$ when these are non-collinear (right).

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Example

By the empty-sum convention, $\mathrm{Vect}(\emptyset) = \{0_E\}$ for every $\mathbb{K}$-vector space $E$.

Example

In $\mathbb{R}^2$, $\mathrm{Vect}((1, 2)) = \{\lambda (1, 2) : \lambda \in \mathbb{R}\}$ is the line through the origin directed by $(1, 2)$. More generally, for any non-zero vector $u$ of $E$, $\mathrm{Vect}(u)$ is the line through the origin $\mathbb{K} u = \{\lambda u : \lambda \in \mathbb{K}\}$.

Example

$\mathbb{K}[X] = \mathrm{Vect}\bigl((X^k)_{k \in \mathbb{N}}\bigr)$: every polynomial is a finite linear combination of the monomials $1, X, X^2, \ldots$ For each $n \in \mathbb{N}$, $\mathbb{K}_n[X] = \mathrm{Vect}(1, X, X^2, \ldots, X^n)$: every polynomial of degree at most $n$ is a linear combination of the first $n+1$ monomials.

Proposition — Properties of Vect

Let $E$ be a $\mathbb{K}$-vector space and $X, Y$ two subsets of $E$, $x \in E$.

Inclusion: if $X \subset Y$, then $\mathrm{Vect}(X) \subset \mathrm{Vect}(Y)$.
Removing a vector: if $x \in X$ is a linear combination of $X \setminus \{x\}$, then $\mathrm{Vect}(X) = \mathrm{Vect}(X \setminus \{x\})$.
Replacing a vector: if $b \in E$ is a linear combination of $X \cup \{a\}$ with a non-zero coefficient on $a$, then $\mathrm{Vect}(X \cup \{a\}) = \mathrm{Vect}(X \cup \{b\})$.

Proof

Inclusion. If $X \subset Y$, every linear combination of elements of $X$ is also a linear combination of elements of $Y$. Hence $\mathrm{Vect}(X) \subset \mathrm{Vect}(Y)$.
Removing a vector. Since $X \setminus \{x\} \subset X$, point 1 gives $\mathrm{Vect}(X \setminus \{x\}) \subset \mathrm{Vect}(X)$. Conversely, by hypothesis $x \in \mathrm{Vect}(X \setminus \{x\})$, and every other element of $X$ also belongs to $\mathrm{Vect}(X \setminus \{x\})$ (which is a subspace containing $X \setminus \{x\}$). Hence $X \subset \mathrm{Vect}(X \setminus \{x\})$, so $\mathrm{Vect}(X) \subset \mathrm{Vect}(X \setminus \{x\})$.
Replacing a vector. Write $b = \lambda a + u$ with $\lambda \ne 0$ and $u \in \mathrm{Vect}(X)$. Then $a = \lambda^{-1}(b - u) \in \mathrm{Vect}(X \cup \{b\})$. Hence $X \cup \{a\} \subset \mathrm{Vect}(X \cup \{b\})$, so $\mathrm{Vect}(X \cup \{a\}) \subset \mathrm{Vect}(X \cup \{b\})$. The reverse inclusion follows from $b \in \mathrm{Vect}(X \cup \{a\})$ (by hypothesis). Hence equality.

Method

To show $F$ is a subspace, often the cleanest path is to write $F$ as a $\mathrm{Vect}$. The proposition guarantees $\mathrm{Vect}(X)$ is a subspace for any subset $X$, so once $F = \mathrm{Vect}(X)$ is exhibited, no further verification is needed. Example: $F = \{(x, y, z) \in \mathbb{R}^3 : x + y + z = 0\} = \{(x, y, -x-y) : x, y \in \mathbb{R}\} = \mathrm{Vect}((1, 0, -1), (0, 1, -1))$, hence a subspace.

Skills to practice

Working with Vect

III Families of vectors

III.1 Generating families

A generating family of $E$ is a family of vectors that spans the whole space: every vector of $E$ is a linear combination of the family. The shortest characterization: $E = \mathrm{Vect}(\text{the family})$.

Definition — Generating family

Let $E$ be a $\mathbb{K}$-vector space and $(x_i)_{i \in I}$ a family of vectors of $E$. The family is generating for $E$ (or generates $E$) if every vector of $E$ is a linear combination of $(x_i)_{i \in I}$, i.e.\ $E = \mathrm{Vect}\bigl((x_i)_{i \in I}\bigr)$.

Example

The family $\bigl((1, 0), (0, 1)\bigr)$ generates $\mathbb{R}^2$: for every $(x, y) \in \mathbb{R}^2$, $(x, y) = x (1, 0) + y (0, 1)$. More generally, for $n \in \mathbb{N}^*$, define $e_i := (0, \ldots, 0, 1, 0, \ldots, 0)$ ($1$ in position $i$). The family $(e_1, \ldots, e_n)$ generates $\mathbb{K}^n$: every $(x_1, \ldots, x_n) \in \mathbb{K}^n$ equals $\sum_{i=1}^n x_i e_i$.

Example

For $i \in \llbracket 1, n \rrbracket$ and $j \in \llbracket 1, p \rrbracket$, denote by $E_{ij} \in M_{n,p}(\mathbb{K})$ the matrix with $1$ in position $(i, j)$ and $0$ elsewhere. The family $(E_{ij})_{1 \le i \le n, 1 \le j \le p}$ generates $M_{n,p}(\mathbb{K})$: every $A = (a_{ij}) \in M_{n,p}(\mathbb{K})$ equals $\sum_{i, j} a_{ij} E_{ij}$.

Method

To find a generating family of a subspace $F$: try to write $F$ as a $\mathrm{Vect}$ of explicit vectors. Concretely, parametrize $F$ by free unknowns (after Gauss reduction if $F$ is defined by linear equations), then factor: $F = \{x_1 u_1 + \dots + x_k u_k : x_1, \ldots, x_k \in \mathbb{K}\} = \mathrm{Vect}(u_1, \ldots, u_k)$, and $(u_1, \ldots, u_k)$ is generating.

Skills to practice

Finding a generating family

III.2 Free and dependent families

A free family is one in which no vector is « redundant »: no vector of the family is a linear combination of the others. Equivalently, the coefficients of a linear combination are uniquely determined by the resulting vector --- the identification trap discussed in the « Identification trap » remark above disappears.

Definition — Free family --- finite case

Let $E$ be a $\mathbb{K}$-vector space and $x_1, \ldots, x_n \in E$. The family $(x_1, \ldots, x_n)$ is free (or the vectors $x_1, \ldots, x_n$ are linearly independent) if $$ \forall (\lambda_1, \ldots, \lambda_n) \in \mathbb{K}^n, \quad \left( \sum_{k=1}^n \lambda_k x_k = 0_E \implies \forall k, \ \lambda_k = 0 \right). $$ Equivalently, the coefficients of a linear combination are unique: if $\sum_k \lambda_k x_k = \sum_k \mu_k x_k$, then $\lambda_k = \mu_k$ for all $k$ (identification of coefficients).

Definition — Dependent family

A family of vectors that is not free is called dependent (or the vectors are linearly dependent). Equivalently, a family is dependent iff one of its terms can be written as a linear combination of the other terms.

Example

The family $(1, i)$ is free in $\mathbb{C}$ viewed as a $\mathbb{R}$-vector space but dependent in $\mathbb{C}$ viewed as a $\mathbb{C}$-vector space.

Answer

In $\mathbb{C}$ as an $\mathbb{R}$-vector space. If $\lambda \cdot 1 + \mu \cdot i = 0$ with $\lambda, \mu \in \mathbb{R}$, then $\lambda + i \mu = 0$, and identifying real and imaginary parts: $\lambda = 0$ and $\mu = 0$. The family $(1, i)$ is free.
In $\mathbb{C}$ as a $\mathbb{C}$-vector space. The coefficient on $1$ can be $-i$: $(-i) \cdot 1 + 1 \cdot i = 0$, with non-zero scalars. Hence the family $(1, i)$ is dependent (two vectors, the second a scalar multiple of the first: $i = i \cdot 1$).

Example

The family $(e_1, \ldots, e_n)$ of $\mathbb{K}^n$ (where $e_i$ has $1$ in position $i$ and $0$ elsewhere) is free in $\mathbb{K}^n$: if $\sum \lambda_i e_i = (\lambda_1, \ldots, \lambda_n) = 0_{\mathbb{K}^n}$, then $\lambda_i = 0$ for all $i$ (identification entry by entry).

Example

The family $\bigl((2, 1), (-1, 3), (0, 1)\bigr)$ is dependent in $\mathbb{R}^2$.

Answer

Search $\lambda, \mu, \nu \in \mathbb{R}$ not all zero with $\lambda (2, 1) + \mu (-1, 3) + \nu (0, 1) = (0, 0)$: $$ \begin{cases} 2 \lambda - \mu = 0 \\ \lambda + 3 \mu + \nu = 0. \end{cases} $$ The first gives $\mu = 2 \lambda$, the second then gives $\nu = -7 \lambda$. With $\lambda = 1$: $(\lambda, \mu, \nu) = (1, 2, -7)$, all non-zero on the first two. Hence the family is dependent.

Proposition — Free family of polynomials of staggered degree

Let $P_1, \ldots, P_n \in \mathbb{K}[X]$ be non-zero polynomials with strictly increasing degrees: $0 \le \deg P_1 < \deg P_2 < \cdots < \deg P_n$. Then the family $(P_1, \ldots, P_n)$ is free.

Proof

Set $d_k := \deg P_k$, and write $P_k = a_{d_k, k} X^{d_k} + (\text{terms of strictly smaller degree})$ with leading coefficient $a_{d_k, k} \ne 0$. Suppose $\sum_{k=1}^n \lambda_k P_k = 0$ for some scalars $\lambda_1, \ldots, \lambda_n \in \mathbb{K}$. By identifying the coefficient of $X^{d_n}$ in the sum: $\lambda_n a_{d_n, n} = 0$ (only $P_n$ contributes to this coefficient, since all other $P_k$ have degree $< d_n$). Since $a_{d_n, n} \ne 0$, $\lambda_n = 0$. The same argument applied to $X^{d_{n-1}}$ gives $\lambda_{n-1} = 0$, and so on down to $\lambda_1 = 0$. Hence all $\lambda_k$ are zero: the family is free.

Example

The family $(1, X, X^2 + 1, X^3 - X)$ has strictly increasing degrees $(0, 1, 2, 3)$, so it is free in $\mathbb{R}[X]$ by the previous proposition.

Method

To show a family $(x_1, \ldots, x_n)$ is free: set $\sum \lambda_k x_k = 0_E$, which translates by identification (entry by entry in $\mathbb{K}^n$, coefficient by coefficient in $\mathbb{K}[X]$, entry by entry in $M_{n,p}(\mathbb{K})$, etc.) into a homogeneous linear system in the unknowns $\lambda_1, \ldots, \lambda_n$. The family is free iff the system has only the trivial solution.

Proposition — Properties of free and dependent families

Let $E$ be a $\mathbb{K}$-vector space and $(x_i)_{i \in I}$ a family of vectors of $E$.

Subfamily of a free family is free. If $(x_i)_{i \in I}$ is free and $J \subset I$, then the subfamily $(x_j)_{j \in J}$ is free. Equivalently, any family containing a dependent subfamily is dependent.
Adding a vector to a free family. If $(x_i)_{i \in I}$ is free and $y \in E$ is not a linear combination of $(x_i)_{i \in I}$, then the family $(x_i)_{i \in I}$ extended by $y$ is free.

Proof

Point 1. Take an almost-zero family $(\lambda_j)_{j \in J}$ of $\mathbb{K}$ with $\sum_{j \in J} \lambda_j x_j = 0_E$. Extend it to an almost-zero family on $I$ by setting $\lambda_i := 0$ for $i \in I \setminus J$. The relation reads $\sum_{i \in I} \lambda_i x_i = 0_E$, hence $\lambda_i = 0$ for all $i \in I$ by freeness of $(x_i)_{i \in I}$. In particular, $\lambda_j = 0$ for all $j \in J$. So $(x_j)_{j \in J}$ is free.
Point 2. Suppose $\sum_{i \in I} \lambda_i x_i + \mu y = 0_E$ for some almost-zero $(\lambda_i)$ and some $\mu \in \mathbb{K}$. If $\mu \ne 0$, then $y = -\mu^{-1} \sum_i \lambda_i x_i \in \mathrm{Vect}\bigl((x_i)_{i \in I}\bigr)$, contradicting the hypothesis on $y$. Hence $\mu = 0$, then $\sum_i \lambda_i x_i = 0_E$, and freeness of $(x_i)_{i \in I}$ gives $\lambda_i = 0$ for all $i$. So the extended family is free.

Skills to practice

Showing a family is free

III.3 Bases and coordinates

A basis combines the two structural properties: generating (existence of a decomposition) and free (uniqueness). Together, every vector of $E$ admits a unique decomposition as a linear combination of the basis vectors. The coefficients in this unique decomposition are the coordinates of the vector in the basis.
This chapter introduces bases in their general form (any indexing set $I$, possibly infinite). The existence of a finite basis, the dimension, the theorem of the incomplete basis and the rank are deferred to the next chapter, Finite-dimensional vector spaces.

Definition — Basis and coordinates

Let $E$ be a $\mathbb{K}$-vector space and $\mathcal{B} = (e_i)_{i \in I}$ a family of vectors of $E$. The family $\mathcal{B}$ is a basis of $E$ if it is both free and generating, equivalently if every vector of $E$ admits a unique expression as a linear combination of $\mathcal{B}$: $$ \forall x \in E, \quad \exists ! \, (\lambda_i)_{i \in I} \text{ almost-zero family of } \mathbb{K}, \quad x = \sum_{i \in I} \lambda_i e_i. $$ The unique family $(\lambda_i)_{i \in I}$ is called the family of coordinates of $x$ in the basis $\mathcal{B}$.

Example

In $\mathbb{R}^2$, the family $\bigl((1, 0), (0, 1)\bigr)$ is a basis: it is generating (every $(x, y) \in \mathbb{R}^2$ writes $(x, y) = x \cdot (1, 0) + y \cdot (0, 1)$) and free (if $\lambda \cdot (1, 0) + \mu \cdot (0, 1) = (0, 0)$, then $(\lambda, \mu) = (0, 0)$). The unique decomposition of $(x, y)$ exhibits its coordinates in this basis: $(x, y)$ itself.

Definition — Canonical bases

The following families are called the canonical bases of their respective spaces:

In $\mathbb{K}^n$: $(e_1, \ldots, e_n)$ where $e_i$ has $1$ in position $i$ and $0$ elsewhere.
In $\mathbb{K}[X]$: the infinite family $(X^k)_{k \in \mathbb{N}} = (1, X, X^2, X^3, \ldots)$.
In $\mathbb{K}_n[X]$: the finite family $(1, X, X^2, \ldots, X^n)$.
In $M_{n,p}(\mathbb{K})$: $(E_{ij})_{1 \le i \le n, \, 1 \le j \le p}$ where $E_{ij}$ has $1$ in position $(i, j)$ and $0$ elsewhere.

Each of these is both generating (already shown) and free (by identification of entries / coefficients).

Example

In the canonical basis of $\mathbb{R}^2$, the coordinates of $(3, -2)$ are $(3, -2)$ themselves: $(3, -2) = 3 e_1 + (-2) e_2$.

Example

The polynomial $P = 2 X^2 - X + 5$ has coordinates $(5, -1, 2)$ in the canonical basis $(1, X, X^2)$ of $\mathbb{R}_2[X]$. In the reordered basis $(X^2, X, 1)$, its coordinates are $(2, -1, 5)$. Coordinates depend on the order of the basis vectors.

Method

To find a basis of a subspace $F$: find a generating family of $F$ (write $F$ as a $\mathrm{Vect}$), then verify the family is free (homogeneous linear system). If the family obtained is free, it is a basis; if not, remove a redundant vector (Proposition « Properties of Vect », point 2) and try again.

Skills to practice

Identifying a basis
Finding coordinates

IV Sum of subspaces

IV.1 Sum of two subspaces

The intersection of subspaces is a subspace (as proved above), but the union is not (see the Warning above). The right « union-like » construction is the sum $F + G$, the smallest subspace containing both $F$ and $G$. Geometrically: the sum of two distinct lines through the origin is the plane through the origin they span.

Definition — Sum of subspaces

Let $E$ be a $\mathbb{K}$-vector space and $F_1, \ldots, F_p$ subspaces of $E$. The sum of $F_1, \ldots, F_p$, written $F_1 + \dots + F_p$, is $$ F_1 + \dots + F_p := \{f_1 + \dots + f_p : f_1 \in F_1, \ldots, f_p \in F_p\}. $$ For two subspaces $F$ and $G$, $F + G = \{f + g : f \in F, g \in G\}$.

Proposition — Sum is a subspace

Let $E$ be a $\mathbb{K}$-vector space and $F_1, \ldots, F_p$ subspaces of $E$. The sum $F_1 + \dots + F_p$ is a subspace of $E$; moreover, it is the smallest subspace of $E$ containing $F_1, \ldots, F_p$ (any subspace containing all the $F_i$ also contains their sum).

Proof

Set $S := F_1 + \dots + F_p$. First, $0_E = 0_E + \dots + 0_E \in S$ (each $F_i$ contains $0_E$). Stability under linear combination: given $\lambda \in \mathbb{K}$ and $s = f_1 + \dots + f_p, s' = f_1' + \dots + f_p' \in S$, $$ \lambda s + s' = (\lambda f_1 + f_1') + \dots + (\lambda f_p + f_p'), $$ and each $\lambda f_i + f_i' \in F_i$ (since $F_i$ is a subspace), so $\lambda s + s' \in S$. Hence $S$ is a subspace.
Minimality: let $F$ be a subspace containing all the $F_i$. For any $s = f_1 + \dots + f_p \in S$, each $f_i \in F_i \subset F$, so $s = f_1 + \dots + f_p \in F$ (stability of $F$ under $+$). Hence $S \subset F$.

Two distinct lines $F$ and $G$ through the origin sum to the plane $F + G$ through the origin, which contains both lines.

Example

In $\mathbb{R}^3$, set $F = \mathrm{Vect}((1, 0, 0))$ and $G = \mathrm{Vect}((0, 1, 0))$ (the two coordinate axes). Then $F + G = \{(x, 0, 0) + (0, y, 0) : x, y \in \mathbb{R}\} = \{(x, y, 0) : x, y \in \mathbb{R}\}$, the $xy$-plane in $\mathbb{R}^3$.

Proposition — Generating family of a sum

Let $E$ be a $\mathbb{K}$-vector space, $F_1, \ldots, F_p$ subspaces of $E$, and $X_1, \ldots, X_p$ generating families of $F_1, \ldots, F_p$ respectively. Then $X_1 \cup \cdots \cup X_p$ is a generating family of $F_1 + \dots + F_p$. Equivalently, $$ \mathrm{Vect}(X_1 \cup \cdots \cup X_p) = \mathrm{Vect}(X_1) + \cdots + \mathrm{Vect}(X_p). $$

Proof

$(\subset)$. Every element of $X_i$ belongs to $\mathrm{Vect}(X_i)$, hence to $\mathrm{Vect}(X_1) + \cdots + \mathrm{Vect}(X_p)$. So $X_1 \cup \cdots \cup X_p$ is contained in this sum, which is a subspace. Hence $\mathrm{Vect}(X_1 \cup \cdots \cup X_p) \subset \mathrm{Vect}(X_1) + \cdots + \mathrm{Vect}(X_p)$.
$(\supset)$. Take $w = v_1 + \cdots + v_p$ with $v_i \in \mathrm{Vect}(X_i) = F_i$. Each $v_i$ is a linear combination of $X_i$, hence of $X_1 \cup \cdots \cup X_p$. A sum of linear combinations of $X_1 \cup \cdots \cup X_p$ is again a linear combination of $X_1 \cup \cdots \cup X_p$, so $w \in \mathrm{Vect}(X_1 \cup \cdots \cup X_p)$.

Method

To compute $F + G$ explicitly: find generating families of $F$ and of $G$, concatenate them, then simplify the resulting $\mathrm{Vect}$ (remove redundant vectors via Proposition « Properties of Vect »).

Skills to practice

Computing a sum

IV.2 Direct sum

A sum $F_1 + \dots + F_p$ is « direct » when each vector of the sum admits a unique decomposition. The notion runs in parallel with free families: dependence in a free family / non-uniqueness of decomposition in a direct sum. For two subspaces, direct sum admits a particularly clean characterization: $F \cap G = \{0_E\}$. For three or more subspaces, this pairwise criterion is not sufficient --- one must check the uniqueness of the decomposition of $0_E$ globally.

Definition — Direct sum

Let $E$ be a $\mathbb{K}$-vector space and $F_1, \ldots, F_p$ subspaces of $E$. The sum $F_1 + \dots + F_p$ is direct if every vector of the sum admits a unique decomposition as $f_1 + \dots + f_p$ with $f_i \in F_i$. We then write $F_1 \oplus \dots \oplus F_p$.

Proposition — Characterization by the decomposition of zero

Let $E$ be a $\mathbb{K}$-vector space and $F_1, \ldots, F_p$ subspaces of $E$. The sum $F_1 + \dots + F_p$ is direct if and only if the zero vector admits a unique decomposition: $$ \forall (f_1, \ldots, f_p) \in F_1 \times \dots \times F_p, \quad (f_1 + \dots + f_p = 0_E \implies f_1 = \dots = f_p = 0_E). $$

Proof

$(\Rightarrow)$. If the sum is direct, every vector --- in particular $0_E$ --- has a unique decomposition. Since $0_E = 0_E + \dots + 0_E$ is one decomposition with $0_E \in F_i$ for each $i$, it must be the only one.
$(\Leftarrow)$. Suppose $0_E$ has a unique decomposition. Let $x \in F_1 + \dots + F_p$ with two decompositions $x = f_1 + \dots + f_p = f_1' + \dots + f_p'$ ($f_i, f_i' \in F_i$). Subtracting: $$ 0_E = (f_1 - f_1') + \dots + (f_p - f_p'), $$ with each $f_i - f_i' \in F_i$. By uniqueness for $0_E$, $f_i - f_i' = 0_E$ for each $i$, i.e.\ $f_i = f_i'$. Hence the decomposition of $x$ is unique.

Proposition — Characterization for two subspaces

Let $E$ be a $\mathbb{K}$-vector space and $F, G$ two subspaces of $E$. The sum $F + G$ is direct if and only if $F \cap G = \{0_E\}$.

Proof

$(\Rightarrow)$. Suppose $F + G$ is direct. Always $\{0_E\} \subset F \cap G$ (each subspace contains $0_E$). Conversely, let $x \in F \cap G$. Then $0_E = x + (-x)$ with $x \in F$ and $-x \in G$ (since $G$ contains $-x = (-1) \cdot x$). Also $0_E = 0_E + 0_E$ with $0_E \in F$ and $0_E \in G$. By uniqueness of the decomposition of $0_E$, $x = 0_E$. Hence $F \cap G \subset \{0_E\}$.
$(\Leftarrow)$. Suppose $F \cap G = \{0_E\}$. Let $f + g = 0_E$ with $f \in F$, $g \in G$. Then $f = -g$, with $f \in F$ and $-g \in G$ (since $G$ is a subspace), so $f \in F \cap G = \{0_E\}$, hence $f = 0_E$, and then $g = -f = 0_E$. By the previous proposition, $F + G$ is direct.

Warning --- pairwise trivial intersection is not enough

For three or more subspaces, pairwise intersection trivial does not imply direct sum. Counter-example in $\mathbb{R}^2$: take $F_1 = \mathrm{Vect}((1, 0))$, $F_2 = \mathrm{Vect}((0, 1))$, $F_3 = \mathrm{Vect}((1, 1))$ (three distinct lines through the origin). Pairwise: $F_1 \cap F_2 = F_1 \cap F_3 = F_2 \cap F_3 = \{(0, 0)\}$. But the sum is not direct, because $(0, 0)$ admits two decompositions: $$ (0, 0) = (0, 0) + (0, 0) + (0, 0) = (1, 0) + (0, 1) + (-1, -1), $$ with $(1, 0) \in F_1$, $(0, 1) \in F_2$, $(-1, -1) \in F_3$. To check direct sum for $p \ge 3$, use the previous proposition (uniqueness of the decomposition of $0_E$ globally).

Example

In $\mathbb{R}^3$, set $F = \{(x, y, z) \in \mathbb{R}^3 : x + y + z = 0\}$ (a plane) and $G = \mathrm{Vect}((1, 1, 1))$ (a line). Show that $F$ and $G$ are in direct sum.

Answer

Compute $F \cap G$. A vector of $G$ has the form $\lambda (1, 1, 1) = (\lambda, \lambda, \lambda)$ for some $\lambda \in \mathbb{R}$. It belongs to $F$ iff $\lambda + \lambda + \lambda = 0$, i.e.\ $\lambda = 0$. Hence $F \cap G = \{(0, 0, 0)\}$. By the previous proposition, $F$ and $G$ are in direct sum.

Method

To show $F$ and $G$ are in direct sum (two subspaces): show $F \cap G = \{0_E\}$. Concretely, take $x \in F \cap G$ and derive $x = 0_E$ (using the defining conditions of $F$ and of $G$).
For $p \ge 3$ subspaces: show the decomposition of $0_E$ is unique --- take $f_1 + \dots + f_p = 0_E$ with $f_i \in F_i$, derive $f_i = 0_E$ for each $i$.

Skills to practice

Recognizing a direct sum

IV.3 Supplementary subspaces

Two subspaces are supplementary in $E$ when they are in direct sum and their sum equals $E$. Equivalently, every vector of $E$ admits a unique decomposition as $f + g$ with $f \in F$, $g \in G$. We strongly encourage the student to picture supplementary spaces with a figure in dimension 2 and 3. The figure that follows the definition is the canonical picture: a plane $F$ in $\mathbb{R}^3$, a line $G$ not contained in $F$, and a generic vector $w$ of $\mathbb{R}^3$ decomposed as $w = f + g$ with $f \in F$ along the plane and $g \in G$ along the line.

Definition — Supplementary subspaces

Let $E$ be a $\mathbb{K}$-vector space and $F, G$ two subspaces of $E$. We say $F$ and $G$ are supplementary in $E$ if $E = F \oplus G$, equivalently if $E = F + G$ and $F \cap G = \{0_E\}$, equivalently if every vector of $E$ admits a unique decomposition $w = f + g$ with $f \in F$ and $g \in G$. We then say $G$ is a supplement of $F$ in $E$ (and vice versa).

A plane $F$ and a transverse line $G$ in $\mathbb{R}^3$: every vector $w$ admits a unique decomposition $w = f + g$ with $f \in F$ and $g$ along $G$.

Warning --- « a supplement » not « the supplement »

There is in general no uniqueness of the supplement. For instance, in $\mathbb{R}^2$, any line through the origin $G \ne F$ is a supplement of a given line $F$ --- there are infinitely many. Hence we always write « $G$ is a supplement of $F$ », never « $G$ is the supplement of $F$ ».
Also do not confuse « supplement » with « complement ». The set-theoretic complement of $F$ in $E$ is $E \setminus F$, which is almost never a subspace (it does not contain $0_E$).

Example

In $\mathbb{R}^3$, the plane $F = \{(x, y, z) \in \mathbb{R}^3 : x - y - z = 0\}$ and the line $G = \{(x, y, z) \in \mathbb{R}^3 : y = z = 0\} = \mathrm{Vect}((1, 0, 0))$ are supplementary.

Answer

Let $(x, y, z) \in \mathbb{R}^3$. We show that $(x, y, z)$ admits a unique decomposition $w = f + g$ with $f \in F$ and $g \in G$.

Analysis. Suppose $(x, y, z) = (x_1, y_1, z_1) + (x_2, y_2, z_2)$ with $(x_1, y_1, z_1) \in F$ (so $x_1 = y_1 + z_1$) and $(x_2, y_2, z_2) \in G$ (so $y_2 = z_2 = 0$). Component identification gives $y = y_1$, $z = z_1$, and $x_1 + x_2 = x$ with $x_1 = y_1 + z_1 = y + z$. Hence $x_1 = y + z$, $y_1 = y$, $z_1 = z$ and $x_2 = x - y - z$, $y_2 = z_2 = 0$. The decomposition (if it exists) is unique.
Synthesis. Set $f := (y + z, y, z)$ and $g := (x - y - z, 0, 0)$. Then $f \in F$ since $(y + z) - y - z = 0$; $g \in G$ since its 2nd and 3rd entries are zero. And $f + g = (y + z + x - y - z, y, z) = (x, y, z)$. Hence the decomposition exists.

Existence and uniqueness $\Rightarrow F \oplus G = \mathbb{R}^3$.

Example

The subspaces $\mathcal{S}_n(\mathbb{R}) := \{S \in M_n(\mathbb{R}) : S^\top = S\}$ (symmetric matrices) and $\mathcal{A}_n(\mathbb{R}) := \{A \in M_n(\mathbb{R}) : A^\top = -A\}$ (antisymmetric matrices) are supplementary in $M_n(\mathbb{R})$.

Answer

Let $M \in M_n(\mathbb{R})$. We show $M$ admits a unique decomposition $M = S + A$ with $S \in \mathcal{S}_n$, $A \in \mathcal{A}_n$.

Analysis. If $M = S + A$ then $M^\top = S^\top + A^\top = S - A$. Half-sum and half-difference yield $S = \tfrac{M + M^\top}{2}$ and $A = \tfrac{M - M^\top}{2}$. The decomposition (if it exists) is unique.
Synthesis. Set $S := \tfrac{M + M^\top}{2}$ and $A := \tfrac{M - M^\top}{2}$. Then $M = S + A$, $S^\top = \tfrac{M^\top + M}{2} = S$ (symmetric), and $A^\top = \tfrac{M^\top - M}{2} = -A$ (antisymmetric). Hence the decomposition exists.

Existence and uniqueness $\Rightarrow \mathcal{S}_n(\mathbb{R}) \oplus \mathcal{A}_n(\mathbb{R}) = M_n(\mathbb{R})$.

Example

The set $\mathcal{P}$ of even functions $\mathbb{R} \to \mathbb{R}$ (i.e.\ $f(-x) = f(x)$) and the set $\mathcal{I}$ of odd functions $\mathbb{R} \to \mathbb{R}$ (i.e.\ $f(-x) = -f(x)$) are supplementary in $\mathbb{R}^\mathbb{R}$: every function $f$ admits a unique decomposition $f = p + i$ with $p$ even and $i$ odd, namely $p(x) = \tfrac{f(x) + f(-x)}{2}$ and $i(x) = \tfrac{f(x) - f(-x)}{2}$ (analysis/synthesis, as for symmetric/antisymmetric matrices).

Method

To show $F$ and $G$ are supplementary in $E$: typically use analysis/synthesis. Take a generic $w \in E$. In the analysis step, assume $w = f + g$ with $f \in F$, $g \in G$, and derive explicit formulas for $f$ and $g$ in terms of $w$ (uniqueness). In the synthesis step, define $f$ and $g$ by those formulas, check $f \in F$, $g \in G$ and $f + g = w$ (existence). Together, $E = F \oplus G$.
Alternative: show separately that $E = F + G$ (every $w$ writes as $f + g$) and $F \cap G = \{0_E\}$.

Skills to practice

Showing supplementary
Finding a supplementary