## Vector Spaces

### Notes on fields

A closer look at the fields $$\Zn{2}[x]/(x^2+x+1)$$

and

$\Zn{3}[x]/(x^2+1)$

with 4 and 9 elements respectively.

### Vector spaces

If $V$ is an abelian group, then let

$\End(V)=\lbrace f: V\to V \hbox{\ where f is a homomorphism}\rbrace$

Proposition: $\End(V)$ is a ring with unity, where:

• addition is addition of maps $(f+g)(v)=f(v)+g(v)$.
• multiplication is composition of maps $(fg)(v)=f(g(v))$.
• The identity map is the identity element for multiplication.
• The zero map is the identity element for addition.

Let $F$ be a field. A non-trivial ring homomorphism (sending $1$ to $1$) $F\to \End(V)$ makes $V$ an $F$ vector space; and an $F$-vector space structure on an abelian group $V$ is equivalent to a non-trivial homomorphism $F\to\End(V)$.

Notice that $\Zn{3}$ maps into $\End(\Zn{6})=\Zn{6}$, but this map doesn’t send $1$ to $1$. So $\Zn{6}$ is not a vector space over $\Zn{3}$.

A linear map $f:V\to W$ is a group homomorphism such that $f(av)=af(v)$ for all $a\in F$. The space $\Hom(V,W)$ of linear maps from $V$ to $W$ is a vector space over $F$. The space $\Hom(V,V)$ of linear maps from $V$ to $V$ is a ring.

Lemma: If $f:V\to V$ is linear and bijective, then its inverse is also linear.

Proof: Let $g=f^{-1}$. Then $g(f(ax)))=ax$ so $g(af(x))=ax$. Write $f(x)=y$ and $x=g(y)$, and we have $g(ay)=ag(y)$.

An isomorphism of vector spaces is a bijective linear map $V\to W$. The units in the ring $\Hom(V,V)$ are the automorphisms of $V$ – that is, the invertible linear maps from $V$ to $V$.

A subspace is a subgroup $W$ such that $aW=W$ for all $a\in F$.

### Basis and Dimension

Definition: A basis of $V$ is a subset that both spans $V$ and is linearly independent.

Proposition: A basis is a minimal spanning set. In other words, if $B$ is a set of vectors that spans $V$, but no proper subset of $B$ spans $V$, then $B$ is a basis.

Proof: Suppose that $B$ is not a basis. Then it is linearly dependent, so there is a finite set of vectors $v_1,\ldots, v_n$ such that $\sum a_{i}v_{i}=0$ with not all $a_i=0$. Therefore we can “solve” for one of the $v_{i}$ in terms of the others, and conclude that there is a proper subset of $B$ that spans $V$.

The ring $F[x]/(f(x))$, where $f(x)$ is a monic polynomial of degree $d$, is a vector space over $F$ with basis $1,x,\ldots, x^{d-1}$.

Corollary: A finite spanning set of $V$ contains a basis.

Proof: Choose a minimal spanning subset.

Proposition: If $A=\lbrace a_1,\ldots, a_n\rbrace$ is a basis for $V$ and $B=\lbrace b_1,\ldots, b_k\rbrace$ is a linearly independent set, then one can reorder the elements of $A$ so that $A’=\lbrace b_1,\ldots, b_k, a_{k+1},\ldots, a_n\rbrace$ is a basis of $V$. In particular, $A$ has at least as many elements as $B$.

Proof: DF give an inductive argument. Axler describes a process for reducing a spanning set to a linearly independent set.
His argument is: put $A$ and $B$ together, with $B$ first: $$b_1,\ldots, b_k, a_1,\ldots, a_n$$ This is a spanning set. The list $b_k, a_1,\ldots, a_n$ must be linearly dependent since $b_k$ is in the span of the $a_i$. This means there’s a linear relation expressing $b_k$ as a sum of $a_i$’s – let’s say $a_n$, renumbering if necessary – so $b_k,a_1,\ldots, a_{n-1}$ is again a basis. Now consider $b_{k-1},b_{k},a_1,\ldots, a_{n-1}$. Again $b_{k-1}$ is a linear combination of $b_{k},a_{1},\ldots$; and this linear combination must involve at least one of the $a_i$ since the $b$’s are linearly independent. So again we can eliminate one of the $a$’s, say $a_{n-1}$ after renumbering, and continue.

Corollary: Suppose $V$ has a basis with $n$ elements. Then any spanning set has at leas $n$ elements, and any independent set has at most $n$ elements.

Corollary: If $V$ has a finite basis, then any two bases have the same number of elements. This number is called the dimension of $V$. If $V$ does not have a finite basis, it is infinite dimensional.

Corollary: Any linearly independent set in a finite dimensional space can be extended to a basis.

Proof: Choose any basis and apply the construction in the proposition above with your given independent set and basis.

Corollary: If $W$ is a subspace of $V$ and $V$ is finite dimensional, then the dimension of $W$ is less than or equal to the dimension of $V$, with equality only when $V=W$.

Proof: Inductively construct a linearly independent set in $W$. The process terminates since it can have at most $\dim(V)$ elements.

Proposition: Any two vector spaces over $F$ of finite dimension $n$ are isomorphic. In particular, any such $V$ is isomorphic to $F^{n}$.