Vector Spaces

\mathbb{F}

The elements of $V$ are called vectors, and the elements of $\mathbb{F}$ are called scalars.

The standard example is

\mathbb{F}^n = \mathbb{F}\times \cdots \times \mathbb{F} = \{(x_1,\ldots,x_n) \mid x_i \in \mathbb{F}\}.

For $a \in \mathbb{F}^n$ , we write its coordinates as $a=(a_1,\ldots,a_n)$ .

Subspaces

U \subseteq V

Equivalently, a nonempty subset $U \subseteq V$ is a subspace if

u,v \in U,\ a,b \in \mathbb{F} \quad\Longrightarrow\quad au+bv \in U.

This single closure condition is usually the most efficient way to check that a subset is a subspace.

Sums and Direct Sums

Let $V_1,\ldots,V_n \subseteq V$ be subspaces. Their sum is

V_1+\cdots+V_n := \{v_1+\cdots+v_n \mid v_i \in V_i,\ i=1,\ldots,n\}.

If for every $i \in \{1,\ldots,n\}$ ,

V_i \cap \left(V_1+\cdots+V_{i-1}+V_{i+1}+\cdots+V_n\right) = \{0\},

then this sum is called a direct sum, and we write

V_1 \oplus \cdots \oplus V_n = V_1+\cdots+V_n.

V_1+\cdots+V_n

Proof.

Clearly $V_1+\cdots+V_n$ is a subspace and contains every $V_i$ .

Now let $U \subseteq V$ be any subspace containing $V_1,\ldots,V_n$ . If $v_i \in V_i$ for all $i$ , then each $v_i \in U$ . Since $U$ is closed under addition,

v_1+\cdots+v_n \in U.

Thus $V_1+\cdots+V_n \subseteq U$ , so it is contained in every subspace containing all the $V_i$ .

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v \in V_1 \oplus \cdots \oplus V_n

Proof.

Suppose

v_1+\cdots+v_n = u_1+\cdots+u_n, \qquad u_i,v_i \in V_i.

Then

(u_1-v_1)+\cdots+(u_n-v_n)=0.

Fix $i$ . Rearranging gives

u_i-v_i = -\sum_{j\neq i}(u_j-v_j).

The left side lies in $V_i$ , while the right side lies in $\sum_{j\neq i}V_j$ . Since the sum is direct,

V_i \cap \sum_{j\neq i}V_j = \{0\}.

Hence $u_i-v_i=0$ . This holds for every $i$ , so $u_i=v_i$ for all $i$ .

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Span and Linear Independence

Let $V$ be a vector space and let $v_1,\ldots,v_n \in V$ . The span of these vectors is

\operatorname{span}(v_1,\ldots,v_n) = \{a_1v_1+\cdots+a_nv_n \mid a_i \in \mathbb{F}\}.

The vectors $v_1,\ldots,v_n$ are linearly independent if

a_1v_1+\cdots+a_nv_n=0 \quad\Longrightarrow\quad a_1=\cdots=a_n=0.

They are linearly dependent if there is a nontrivial relation

a_1v_1+\cdots+a_nv_n=0

with at least one coefficient $a_i$ nonzero.

\beta=\{v_1,\ldots,v_n\}

More generally, a subset $U \subseteq V$ is linearly independent if every finite subset of $U$ is linearly independent. It is a spanning set of $V$ if every vector $v \in V$ can be written as a finite linear combination of elements of $U$ .

For arbitrary vector spaces, bases need not be finite. Such bases are often called Hamel bases.

Zorn's Lemma and Hamel Bases

(\mathcal{P},\leq)

Zorn's lemma is one of the standard forms of the axiom of choice. It is exactly the kind of tool that lets us prove existence theorems where no explicit construction is available.

V

Proof.

Let

\mathcal{P} = \{S \subseteq V \mid S \text{ is linearly independent}\},

ordered by inclusion. Since $\varnothing$ is linearly independent, $\mathcal{P}$ is nonempty.

Let $\mathcal{C}\subseteq\mathcal{P}$ be a chain, and define

P=\bigcup_{S\in\mathcal{C}}S.

We claim that $P$ is linearly independent. Indeed, take finitely many vectors $u_1,\ldots,u_m \in P$ and suppose

a_1u_1+\cdots+a_mu_m=0.

For each $i$ , choose $S_i \in \mathcal{C}$ such that $u_i \in S_i$ . Since $\mathcal{C}$ is a chain, among the finitely many $S_i$ there is one that contains all the others. Hence all $u_i$ lie in a single linearly independent set from the chain, so $a_1=\cdots=a_m=0$ .

Thus $P\in\mathcal{P}$ and $P$ is an upper bound for $\mathcal{C}$ . By Zorn's lemma, $\mathcal{P}$ has a maximal element $\mathcal{M}$ .

We now show that $\operatorname{span}(\mathcal{M})=V$ . Suppose not. Choose

v \in V \setminus \operatorname{span}(\mathcal{M}).

Then $\mathcal{M}\cup\{v\}$ is linearly independent: if

av+a_1u_1+\cdots+a_mu_m=0, \qquad u_i\in\mathcal{M},

and $a\neq 0$ , then

v=-\frac{a_1}{a}u_1-\cdots-\frac{a_m}{a}u_m,

which would put $v$ in $\operatorname{span}(\mathcal{M})$ , a contradiction. Therefore $a=0$ , and the independence of $\mathcal{M}$ forces $a_1=\cdots=a_m=0$ .

So $\mathcal{M}\cup\{v\}$ is a strictly larger linearly independent set, contradicting the maximality of $\mathcal{M}$ . Hence $\mathcal{M}$ spans $V$ , and therefore it is a basis of $V$ .

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Exchange and Complements

V

Proof.

Consider the partially ordered set

\mathcal{P} = \{S\subseteq W \mid U\cup S \text{ is linearly independent}\},

ordered by inclusion. By the same chain-union argument used in Hamel's theorem, every chain in $\mathcal{P}$ has an upper bound in $\mathcal{P}$ . Zorn's lemma gives a maximal element $W'\subseteq W$ .

We claim $U\cup W'$ spans $V$ . If not, since $W$ spans $V$ , there must be some $w\in W$ not lying in $\operatorname{span}(U\cup W')$ . Then

U\cup W'\cup\{w\}

is still linearly independent, contradicting the maximality of $W'$ . Therefore $U\cup W'$ is both linearly independent and spanning, so it is a basis of $V$ .

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V

Proof.

Choose a basis $\{u_1,\ldots,u_k\}$ of $U$ . Since this set is linearly independent in $V$ , the exchange lemma extends it to a basis

\{u_1,\ldots,u_k,w_1,\ldots,w_m\}

of $V$ . Define

W=\operatorname{span}(w_1,\ldots,w_m).

Every vector in $V$ is a sum of something in $U$ and something in $W$ , so $U+W=V$ . The intersection $U\cap W$ is trivial because the displayed basis is linearly independent. Hence

U\oplus W=V.

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