# Finite-Dimensional Linear Spaces

Do you know what linearly independent means? Do you know what a spanning set is? A vector space with a finite linearly independent set of spanning vectors is called a finite-dimensional linear space. In this article, I go over the basic theorems for these spaces by providing their proofs.

We study finite-dimensional linear spaces in the abstract. For background reading you might want to read about subspaces and linear independence first.

## Direct Sums of Subspaces of Finite-Dimensional Linear Spaces

Definition. A vector space $\mathbb{V}$ is a direct sum of subspaces $U_1, \ldots, U_m$ of $\mathbb{V}$, written $V=U_{1} \oplus \cdots \oplus U_{m}$, if each element of $\mathbb{V}$ can be written uniquely as a sum $u_1 + u_2+\cdots + u_m$ where each $u_j\in U_j.$

Theorem. If $U_1, \ldots, U_m$are subspaces of $\mathbb{V}$, then $V=U_{1} \oplus \cdots \oplus U_{m}$ if and only if both conditions hold:

(1) $V=U_{1} + \cdots + U_{m}.$

(2) the only way to write $0$ as a sum $u_{1} + u_{2}+\cdots+ u_{m}$ where each $u_{j} \in U_{j}$, is by taking all the $u_{j}$’s equal to $0.$

Proof. If $V=U_{1} \oplus U_2 \oplus \cdots \oplus U_{m}$, then every element in $\mathbb{V}$ can be written uniquely in the form $u_1+ u_2+\cdots +u_m$ where each $u_j\in U_j$ for $1\leq i \leq m.$ Thus both conditions listed above are satisfied. Conversely, suppose both conditions hold and assume $u=u_1 + \cdots +u_m$ where $u_i \in U_i$ and $u=v_1 + \cdots + v_m$ where $v_i\in U_i.$ Since \begin{align*} u-u & =(u_1 + \cdots +u_m)-(v_1 + \cdots +v_m) \\ & =(u_1-v_1)+\cdots +(u_m-v_m) \\ & =0 \end{align*} it follows $u_i=v_i$ for $1\leq i \leq m$; and so uniqueness is established.

Example. Let $\mathbb{V}$ be the linear space of all functions from $\mathbb{R}$ to $\mathbb{R}$ and let $V_e$ and $V_o$ be the set of all even functions $(f(-x)=f(x))$ and the set of odd functions $(f(-x)=-f(x))$, respectively. Prove that $V_e$ and $V_o$ are subspaces of $\mathbb{V}.$ Prove that $V_e+V_o=V$ Prove that $V_e\cap V_o=0.$

Theorem. If $U$ and $W$ are subspaces of $\mathbb{V}$, then $V=U \oplus W$ if and only if $V=U+W$ and $U \cap W= \{0\}.$

Proof. If $V=U \oplus W$ then $V=U+W$ and $0 \subseteq U\cap W$ are immediate. Suppose $v\in U \cap W.$ Since $v\in U \oplus W$, there exists unique $u\in U$ and $w\in W$ such that $v=u+w.$ Assume for a contradiction, $u\not = 0.$ Then $u=u+0$ where $0\in W$ and $u \in U$ shows that $u\not \in W$ since $V=U\oplus W.$ Also, $v=u+w\in U\cap W$ so $u+w\in W$ and $w\in W$ implies $u\in W.$ This contradiction leads to $u=0.$ Similarly, $w=0$ and so $v=u+w=0$ which yields $U \cap W= \{0\}.$ Conversely, suppose $V=U+W$ and $U \cap W= \{0\}.$ If $0=u+w$ then $u=-w\in W$ and together with $u\in U$ yields $u=0.$ Thus, $w=0$ also. So the only way to write 0 as a sum $u+w$ is to have $u=w=0$ therefore $V=U \oplus W.$

Example. Prove or give a counterexample: if $U_1, U_2$, and $W$ are subspaces of $\mathbb{V}$ such that $U_1 +W=U_2+W$, then $U_1=U_2.$ False, here is a counterexample. Let $V=\mathbb{R}^2=W$ with $U_1=\{(x,0) \mid x \in \mathbb{R}\}$ and $U_2=\{(0,y) \mid y\in \mathbb{R}\}.$ Then $U_1+W=U_2+W=\mathbb{R}^2.$ However, $U_1\neq U_2.$

Example. Prove or give a counterexample: if $U_1, U_2, W$ are subspaces of $\mathbb{V}$ such that $V=U_1 \oplus W$ and $V=U_2 \oplus W$, then $U_1=U_2.$ False, here is a counterexample. Let $V=\mathbb{R}^2$ with $W=\{(x,x) \mid x\in \mathbb{R}\}$, $U_1=\{(x,0) \mid x \in \mathbb{R}\}$, and $U_2=\{(0,y) \mid y\in \mathbb{R}\}.$ All of these sets are subspaces. Let $u\in U_1\cap W$, then $u=(x,0)$ and $u=(z,z)$ so $z=0$ and $x=0$ which implies $u=(0,0).$ In fact $U_1\cap W={0}.$ Thus, $\mathbb{R}^2=U_1\oplus W.$ Also $\mathbb{R}^2=U_2\oplus W.$ However, $(1,0)\in U_1$ and $(1,0)\not \in U_2$ showing $U_1\neq U_2.$

Example. Suppose $m$ is a positive integer. Is the set consisting of 0 and all polynomials with coefficients in $F$ and with degree equal to $m$ a subspace of $\mathcal{P}(\mathbb{F})$?

## What is a Finite-Dimensional Linear Spaces?

Definition. We call a linear space $\mathbb{V}$ a finite-dimensional linear space if there is a finite list of vectors $(v_1, \ldots, v_m)$ with $\text{span}(v_1, \ldots, v_m)=V.$ If a linear space is not a finite-dimensional vector space it is called an infinite-dimensional vector space.

Lemma. Let $\mathbb{V}$ be a finite-dimensional vector space. Then the length of every linearly independent list of vectors in $\mathbb{V}$ is less than or equal to the length of every spanning list of vectors in $\mathbb{V}.$

Lemma. Let $\mathbb{V}$ be a finite-dimensional vector space. Then $\mathbb{V}$ only has finite-dimensional subspaces.

## Basic Theorems

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then every spanning list $(v_1,\ldots,v_m)$ of a vector space can be reduced to a basis of the vector space.

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then if $\mathbb{V}$ is finite-dimensional then $\mathbb{V}$ has a basis.

Proof. By definition, every finite-dimensional vector space has a finite spanning set. Every spanning set can be reduced to a basis, and so every finite-dimensional vector space does indeed have a basis.

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then if $W$ is a subspace of $\mathbb{V}$, every linearly independent subset of $W$ is finite and is part of a finite basis for $W.$

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then every linearly independent list of vectors in $\mathbb{V}$ can be extended to a basis of $\mathbb{V}$,

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then if $U$ is a subspace of $\mathbb{V}$, then there is a subspace $W$ of $\mathbb{V}$ such that $V=U \bigoplus W$.

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then any two bases of $\mathbb{V}$ have the same length.

Definition. The dimension of a finite-dimensional vector space is defined to be the length of any basis of the vector space.

## Dimension Theorem for Vector Spaces

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then if $U$ is a subspace of $\mathbb{V}$ then $\mathop{dim} U \leq \mathop{dim} V.$

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then every spanning list of vectors in $\mathbb{V}$ with length dim $\mathbb{V}$ is a basis of $\mathbb{V}$.

Theorem. Let $\mathbb{V}$ be a finite-dimensional vector space. Then every linearly independent list of vectors in $\mathbb{V}$ with length dim $\mathbb{V}$ is a basis of $\mathbb{V}$.

Exercise. Prove that if $U_1$ and $U_2$ are subspaces of a finite-dimensional vector space, then $\mathop{dim} (U_1+U_2)= \mathop{dim} U_1 + \mathop{dim} U_2 -\mathop{dim} (U_1 \cap U_2).$

Theorem. If $\mathbb{V}$ is finite-dimensional, and $U_1, \ldots, U_m$ are subspaces of $\mathbb{V}$ such that

(1) $V=U_1+\cdots+U_m$ and

(2) $\mathop{dim} V=\mathop{dim} U_1+\cdots+\mathop{dim} U_m$,

then $V=U_1 \oplus \cdots \oplus U_m.$

Proof. Choose a basis for each $U_j.$ Put these bases together in one list, forming a list that spans $\mathbb{V}$ and has length $\mathop{dim} (V).$ Thus this list is a basis of $\mathbb{V}$ and in particular it is linearly independent. Now suppose that $u_j\in U_j$ for each $j$ are such that $0=u_1+\cdots + u_m.$ We can write each $u_j$ as a linear combination of the basis vectors of $U_j.$ Substituting these expressions above, we have written $0$ as a linear combination of the basis vectors of $\mathbb{V}.$ Thus all scalars used in the linear combinations must be zero. Thus each $u_j=0$ which proves $V=U_1 \oplus \cdots \oplus U_m.$ David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.