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