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Vector Spaces

Quick reminder about fields

Fields we know about:

  • $\R$, $\Q$, $\C$, $\Q(x)$, … these are fields of characteristic zero
  • $\Zn{p}$ where $p$ is prime, these are finite fields of characteristic $p$.
  • $\Zn{p}(x)$, rational functions with coefficients in $\Zn{p}$, this is an infinite field of characteristic $p$.
  • $\Zn{p}[x]/(f(x))$ where $f(x)$ is an irreducible polynomial of degree $d$ over $\Zn{p}$, this is a finite field with $p^{d}$ elements. For example



Next semester we will prove the following.

Theorem: If $F$ is a finite field of characteristic $p$, then $F$ has $p^{d}$ elements for some $d\ge 1$ and all finite fields of the same order are isomorphic.

Key definitions

In the following, $F$ is a field.

Definition: A vector space $V$ over $F$ is an abelian group together with a map $F\times V\to V$, called scalar multiplication, which satisfies, for all $a,b\in F$ and $v,w\in V$:

  • $a\cdot (b\cdot v)=(ab)\cdot v$
  • $(a+b)\cdot v=a\cdot v + b\cdot v$
  • $a\cdot (v+w)=a\cdot v+a\cdot w$
  • $1\cdot v = v$

Remark: If, in the above definition, we replace $F$ by a ring $R$ with $1$, then the same axioms characterize an object called a left $R$-module. So a module is like a vector space but you only have scalar multiplication by elements of a ring instead of a field.

Definition: Let $V$ be a vector space over $F$.

  • A subspace is a subgroup of $V$ closed under scalar multiplication.
  • A linear map $f:V\to W$ is a group homomorphism that satisfies $f(av)=af(v)$ for all $a\in F$.
  • A (possibly infinite) set $S$ of vectors in $V$ is called linearly independent if, for any finite set $v_1,\ldots, v_k$ of elements of $S$, if $\sum_{i=1}^{k} a_i v_i=0$ then all $a_i=0$.
  • A set of vectors $S$ is said to span $V$ if it generates $V$ as a vector space, meaning the smallest subspace of $V$ containing $S$ is all of $V$.
  • A linearly ordered set of vectors $S$ is a basis of $V$ if it is linearly independent and spans $V$.

Basis and dimension

If a vector space $V$ has a finite basis with $n$ elements, then every basis of $V$ has $n$ elements and $n$ is called the dimension of $V$.

Every vector space of dimension $n$ over $F$ is isomorphic to each other and to $F^{n}$.

The group of bijective linear maps from $V$ to $V$ is called $\Aut(V)$ or $\GL(V)$.


If $F$ is a finite field with $q=p^{d}$ elements, and $W$ is a vector space of dimension $k$, then:

  1. The number of distinct bases of $W$ is $(q^{k}-1)(q^{k}-q)(q^{k}-q^2)\cdots (q^{k}-q^{k-1})$.
  2. The number of subspaces of dimension $k$ is \(\frac{(q^{n}-1)(q^{n}-q)\cdots (q^{n}-q^{k-1})}{(q^{k}-1)(q^{k}-q)\cdots(q^{k}-(q^{k-1})}\)
  3. The group $\Aut(V)$ has the same order as in part 1. (To see this, fix a basis of $V$. Given another basis, there is a bijective linear map from the fixed basis to this new basis. So the number of linear maps is the same as the number of different bases of $V$)

Subspaces and quotients

The kernel of a linear map $f:V\to K$ is a subspace of $V$.

If $W\subset V$ is a subspace, the quotient group $V/W$ is a vector space. It satisfies the “isomorphism theorem” that any linear map $g:V\to K$ such that $W\subset\ker(g)$ factors through the quotient $W/V$:

\[\begin{xy} \xymatrix { V\ar[rd]^{f}\ar[d]^{\pi} & \\ V/W\ar[r]_{\overline{f}} & K\\ } \end{xy}\]

Proposition: If $V$ is finite dimensional, then $\dim(V)=\dim(W)+\dim(V/W)$. (In the infinite dimensional case, both sides are infinite).

Proposition: If $f:V\to W$ is a linear map between vector spaces, then the image $f(V)$ is a subspace of $W$ and $\dim(V)=\dim\ker(f)+\dim f(V)$. This follows from the isomorphism theorem and the preceeding result.

A little Zorn

Theorem: Every vector space has a basis.

Proof: Let $V$ be a vector space and consider the collection of linearly independent subsets of $V$ ordered by inclusion. This is a nonempty set, and if $A_1\subset A_2\subset \ldots$ is a chain, then the union of the $A_i$ is an independent set containing all of the $A_i$. So every ascending chain has an upper bound. By Zorn’s Lemma, the set of linearly independent subsets has a maximal element $B$. Let $x\in W$. The set $B\cap \lbrace x\rbrace$ must be linearly dependent, since $B$ is maximal, so $x$ is a linear combination of elements of $B$. Thus $B$ is a basis of $V$.


Let $V$ and $W$ be finite dimensional vector spaces with basis $A$ and $B$ respectively. Let $f$ be a linear map from $V$ to $W$. Then we have equations

\[f(a_j)=\sum e_{ij}b_{i}\]

for each $j$ between $1$ and $n=\dim(V)$ and $i$ between $1$ and $m=\dim(W)$ respectively. TAKE NOTE OF HOW THE INDICES ARE ORGANIZED


\[M^{B}_{A}(f)=\left[\begin{matrix} e_{11} & e_{12} &\cdots & e_{1n}\\\vdots & \vdots &\cdots & \vdots \\ e_{m1} & e_{m2} & \cdots & e_{mn}\end{matrix}\right]\]

Thus we associate to a linear map $f:V\to W$ an $m\times n$ matrix where $n=\dim(V)$ and $m=\dim(W)$. This depends on the choice of bases $A$ and $B$.

This correspondence has the property that, if $v\in V$ satisfies $v=\sum_{j=1}^{n} x_{j}a_{j}$ then $f(v)=\sum_{j=1}^{m} y_{j}b_{j}$ where

\[M_{A}^{B}(f)\left[\begin{matrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{matrix}\right] = \left[\begin{matrix} y_1 \\ y_2 \\ \vdots \\ y_{m}\end{matrix}\right]\]

Proposition: The map sending $f:V\to W$ to $M_{A}^{B}(f)\in M_{m\times n}(F)$ is an isomorphism of vector spaces between $\Hom(V,W)$ and $M_{m\times n}(F)$.. If $V=W$ and $A=B$, it is a ring isomorphism from $\Hom(V,V)$ to $M_{n}(F)$.