| title |
Vector Spaces |
| author |
Keith A. Lewis |
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relation, cosets,
functions, 1-1, onto, 1-1 correspondence
Let $\FF$ be a field and the set exponential ${\FF^I = {x\colon I\to\FF} = {I\to\FF}}$
be the set of all functions from the set $I$ to $\FF$.
Define scalar multiplication $\FF\times\FF^I\to\FF^I$
where ${(a,x)\mapsto ax = xa}$ by ${(ax)(i) = a(x(i)) = (xa)(i)}$ and vector addition
${\FF^I\times\FF^I\to\FF^I}$ where ${(x,y)\mapsto x + y}$ by
${(x + y)(i) = x(i) + y(i)}$, $i\in I$. We say these are defined pointwise
since $x,y\in\FF^I$ are equal if and only if $x(i) = y(i)$ for all $i\in I$.
Exercise. Show $x + y = y + x$ for $x,y\in\FF^I$.
Hint Evaluate at $i\in I$ and use the fact field addition is commutative.
Solution
We have $(x + y)(i) = x(i) + y(i) = y(i) + x(i) = (y + x)(i)$ for all $i\in I$.
Define $\zero\in\FF^I$ by $\zero(i) = 0$ for $i\in I$. We write
$\zero^I$ to indicate the domain is $\FF^I$. Note
${\zero^I\not=\zero^J}$ if ${I\not=J}$.
Exercise. Show $x + \zero = x$ for all $x\in\FF^I$.
Define $-x\in\FF^I$ by $(-x)(i) = -x(i)$ for $x\in\FF^I$ and $i\in I$.
Exercise. Show $-x = (-1)x$ and $x + (-x) = \zero$ for $x\in\FF^I$.
This shows $\FF^I$ is an abelian group under vector addition with
additive identity $\zero$.
The standard basis of $\FF^I$ is the set of vectors $e_i\colon I\to\FF$
defined by $e_i(j) = 1(i = j)$, $i,j\in I$, where $1(i = j) = 1$ if $i = j$
and $1(i = j) = 0$ if $i\not=j$. This is also called the Kronecker delta $\delta_{ij} = 1(i = j)$.
Exercise. Show $x = \sum{i\in I} x(i) e_i$, $x\in\FF^I$, if $I$ is finite._
Hint. Compute $(\sum_{i\in I} x(i) e_i)(j) = \sum_{i\in I} x(i) e_i(j)$.
Solution
We have $(\sum_{i\in I} x(i) e_i)(j) = \sum_{i\in I} x(i) e_i(j) =
\sum_{i\in I} x(i)1(i = j) = x(j)$.
This shows every $x\in\FF^I$ is a linear combination of the standard basis
vectors.
Exercise. Show if $\sum{i\in I} a_i e_i = \zero$ then $a_i = 0$ for all $i\in I$_.
Hint: Evaluate at $j\in I$.
Any set of vectors with this property is called independent
The dual space of $\FF^I$ is $(\FF^I)^* = \FF^{e_I}$ where
$e_I = {e_i\mid i\in I}$. For $x\in\FF^I$ define $x^\in\FF^{e_I}$
by $x^(e_i) = x(i)$, $i\in I$.
Exercise. Show $(ax)^* = ax^$ and $(x + y)^ = x^* + y^*$ for $a\in\FF$, $x,y\in\FF^I$.
Exercise. Show $x^* = \zero$ implies $x = \zero$ for $x\in\FF^I$
and $(\FF^I)^* = {x^*\mid x\in\FF^I}$.
Any function $T\colon\FF^I\to\FF^J$ satisfying $T(ax) = aTx$ and $T(x + y) = Tx + Ty$ is
called a linear operator.
The set exponential $\FF^I$ can be identifed with the cartesian product of $I$
copies of $\FF$.
Define projections $e_i^\colon\FF^I\to\FF$ by $e_i^(x) = x(i)$.
Exercise. Show if $p_i\colon V\to\FF$, $i\in I$, then there exists
$p\colon V\to\FF^I$ with $p_i = e_i^*p$.
Hint: Define $(pv)(i) = p_i(v)$ and note $(pv)(i) = e_i^*(pv)$.
Vector Space Axioms
Define $\zero\in\FF^I$ by $\zero(i) = 0$ for $i\in I$.
Exercise. Show $0x = \zero$ for $x\in\FF^I$.
Exercise. Show $x + \zero = x = \zero + x$ for all $x\in\FF^I$.
This shows $\zero$ is the identity element for vector addition.
Exercise. Show $x + y = y + x$ for $x, y\in\FF^I$.
Vector addition is commutative.
Exercise. Show $(a + b)x = ax + bx$ and $a(x + y) = ax + ay$ for $a,b\in\FF$ and $x,y\in\FF^I$.
Hint. Consider $(a + b)x(i)$ and $a(x + y)(i)$.
This establishes the distributive laws satisfied by scalar multiplication and vector addition.
Exercise. Show $-x = -1x$ satisfies $x + (-x) = \zero$.
Hint. Consider $x + (-x) = 1x + (-1x) = (1 + (-1))x$.
This shows $-x$ is the additive inverse of $x\in\FF^I$
and vector addition is a commutative group.
Exercise. Show $a(bx) = (ab)x$ for $a,b\in\FF$, $x\in\FF^n$.
Hint: Use the fact multiplication is associative in $\FF$.
Any set $V$ with a scalar multiplication and
vector addition satisfying
the above axioms is a vector space. We will eventually prove
every vector space is isomorphic to $\FF^I$ for some set $I$
but proofs involving only the axioms are more insightful.
A linear operator $T\colon\FF^I\to\FF^J$ is a function
satisfying $T(ax + y) = aT(x) + T(y)$, $a\in\FF$, $x,y\in\FF^I$.
The set of all such operators is denoted $[\FF^I\to\FF^J]$.
It is a vector space with scalar multiplication $(aT)(x) = a(T(x))$
and vector addition $(T + U)(x) = Tx + Ux$, $a\in\FF$, $x\in\FF^I$, $T,U\in[\FF^I\to\FF^J]$.
Since $e_i\in\FF^I$ we have $Te_i = \sum_{j\in I} (Te_i)(j)e_j$
and we define $t(i,j) = Te_i(j)\in\FF$.
If $T\colon\FF^I\to\FF^J$ and $S\colon\FF^J\to\FF^K$ then
$\sigma\colon I\to J$ then $\circ\sigma\colon\FF^J\to\FF^I$ by $x\mapsto x\circ\sigma$.