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\newtheorem*{defn}{Defintion}
\newtheorem*{prop}{Proposition}
\newtheorem*{thm}{Theorem}
\newtheorem*{cor}{Corollary}

\begin{document}
\title{Lecture 2}
\maketitle
\begin{defn}
A subset $W$ of a vector space $V$ is a \underline{subspace} if
\begin{enumerate}
\item $W$ is non-empty
\item For every $\bar{v},\bar{w}\in W$ and $a,b\in\mathbb F$, $a\bar{v}+b\bar{w}\in W$.
\end{enumerate}
\end{defn}
Expressions like $a\bar{v}+b\bar{w}$, or more generally
\[
\sum_{i=1}^k a_i\bar{v}+i
\]
are called \underline{linear combinations}. So a non-empty subset of $V$ is a subspace if it is closed under linear combinations. Much of today's class will focus on properties of subsets and subspaces detected by various conditions on linear combinations.

\begin{thm}
If $W$ is a subspace of $V$, then $W$ is a vector space over $\mathbb F$ with operations coming from those of $V$.
\end{thm}

In particular, since all of those axioms are satisfied for $V$, then they are for $W$. We only have to check closure!

Examples:
\begin{defn}
Let $\mathbb F^n=\{(a_1,\dots,a_n)|a_i\in\mathbb F\}$ with coordinate-wise addition and scalar multiplication.
\end{defn}
This gives us a few examples. Let $W\subset\mathbb F^n$ be those points which are zero except in the first coordinate:
\[
W=\{(a,0,\dots,0)\}\subset\mathbb F^n.
\]
Then $W$ is a subspace, since
\[
a\cdot(\alpha,0,\dots,0)+b\cdot(\beta,0,\dots,0)=(a\alpha+b\beta,0,\dots,0)\in W.
\]

If $\mathbb F=\mathbb R$, then $W'=\{(a_1,\dots,a_n)|a_i\geq 0\}$ is not a subspace. It's closed under addition, but not scalar multiplication.

We have a number of ways to build new subspaces from old.

\begin{prop}
If $W_i$ for $i\in I$ is a collection of subspaces of $V$, then
\[
W=\bigcap_{i\in I}W_i=\{\bar{w}\in V | \bar{w}\in W_i\forall i\in I\}
\]
is a subspace.
\end{prop}
\begin{proof}
Let $\bar{v},\bar{w}\in W$. Then for all $i\in I$, $\bar{v},\bar{w}\in W_i$, by definition. Since each $W_i$ is a subspace, we then learn that for all $a,b\in\mathbb F$,
\[
a\bar{v}+b\bar{w}\in W_i,
\]
and hence $a\bar{v}+b\bar{w}\in W$.
\end{proof}
Thought question: Why is this never empty?

The union is a little trickier.
\begin{prop}
$W_1\cup W_2$ is a subspace iff $W_1\subset W_2$ or $W_2\subset W_1$.
\end{prop}
\begin{proof}
$\Leftarrow$ is obvious. We need to show the other, so assume that we can find $\bar{w}_1\in W_1-W_2$ and $\bar{w}_2\in W_2-W_1$. Then if $W_1\cup W_2$ is a subspace, then $\bar{w}_1+\bar{w}_2\in W_1\cup W_2$. This means it's in one of them, and without loss of generality, we may assume it's in $W_1$. But this means
\[
w_2=(\bar{w}_1+\bar{w}_2)-\bar{w}_1\in W_1,
\]
a contradiction.
\end{proof}

Example to keep in mind: $\mathbb R^2$ with $W_1=x$-axis and $W_2=y$-axis.

Instead of the union, we consider the smallest subspace containing the union.
\begin{defn}
If $W_1$ and $W_2$ are subspaces of $V$, then the \underline{sum} is
\[
W_1+W_2=\{a\bar{w}_1+b\bar{w}_2 | \bar{w}_i\in W_i\}.
\]
\end{defn}
In other words, we consider all linear combinations of elements of $W_1$ and $W_2$. Clearly if $W$ is any subspace that contains $W_1$ and $W_2$, then $W$ contains $W_1+W_2$. On the other hand, it's also closed under linear combinations and non-empty, so this is a subspace.

Special attention is given to the case where $W_1\cap W_2=\{\bar{0}\}$.
\begin{defn}
If $W_1\cap W_2=\{\bar{0}\}$, then we say the sum of $W_1$ and $W_2$ is the (internal) \underline{direct sum}, and we write it $W_1\oplus W_2$.
\end{defn}
So what's so special about direct sums?
\begin{prop}
Every element in $W_1\oplus W_2$ can be uniquely written as $\bar{w}_1+\bar{w}_2$.
\end{prop}
\begin{proof}
Assume $\bar{w}_1+\bar{w}_2=\bar{v}_1+\bar{v}_2$, where the subscript indicates the subspace from which the element is drawn. Then rearranging, we see
\[
\bar{w}_1-\bar{v}_1=\bar{v}_2-\bar{w}_2.
\]
The left-hand side is in $W_1$, while the right-hand side is in $W_2$, so both are in the intersection. This means both are $\bar{0}$, and hence $\bar{w}_i=\bar{v}_i$.
\end{proof}

Our notion of sums therefore includes two distinct notions:
\begin{enumerate}
\item being able to write elements as linear combinations
\item being able to do so uniquely.
\end{enumerate}

The former is related to span, the latter to linear independence.

\begin{defn}
Let $X\subset V$. Then the \underline{span} of $X$ is the collection of all linear combinations of elements of $X$:
\[
Span(X)=\langle X\rangle=\{a_1\bar{x}_1+\dots+a_n\bar{x}_n | \bar{x}_i\in X, n\geq 0\}.
\]
If $X$ is empty, then $\langle X\rangle = \{\bar{0}\}$.
\end{defn}
The following is immediate.
\begin{prop}
The set $\langle X\rangle$ is a subspace of $V$.
\end{prop}

Examples:
For any $V$, $\langle V\rangle=V$.
If $X=W\cup U$, then $\langle X\rangle=W+U$.

Just as before, if $W$ is a subspace of $V$ and $W$ contains $X$, then $\langle X\rangle \subset W$. Thus $\langle X\rangle$ is the smallest subspace containing $X$, and the elements of $X$ provide convenient names for every element of their span.

\begin{prop}
If $\bar{w}\in\langle X\rangle$, then
\[
\langle\{\bar{w}\}\cup X\rangle = \langle X\rangle.
\]
\end{prop}
\begin{proof}
One inclusion is obvious. For the other, since $\bar{w}\in\langle X\rangle$, we know we can write it as a linear combination of elements of $X$. Thus if we have a linear combination of elements of $X$ and $\bar{w}$, then we can substitute for $\bar{w}$ and get a linear combination with just elements of $X$.
\end{proof}

This helps us greatly in cutting out redundant elements: anything we can name using a smaller set of elements is extraneous.
\begin{defn}
A \underline{spanning set} is a set $X$ such that $\langle X\rangle=V$.
\end{defn}
In general, these can be very big. One of the goals of much of linear algebra is to give a very compact spanning set for an arbitrary vector space.

The corresponding small notion is linear independence.
\begin{defn}
A set $X$ is \underline{linearly independent} if
\[
a_1\bar{v}_1+\dots+a_n\bar{v}_n=\bar{0}
\]
implies $a_1=\dots=a_n=0$ for any $\bar{v}_i\in X$.

If $X$ is not linearly independent, then it is linearly dependent.
\end{defn}
We again see only $\bar{0}$ showing up. We can restate this definition as ``$\bar{0}$ has a unique presentation as a linear combination of elements of $X$.

\begin{thm}
If $X$ is linearly independent and
\[
a_1\bar{v}_1+\dots a_n\bar{v}_n=b_1\bar{w}_1+\dots+b_m\bar{w}_m,
\]
where all vectors are from $X$ and all coefficients are non-zero, then $n=m$ and up to reordering, $a_i=b_i$ and $\bar{v}_i=\bar{w}_i$ for all $i$.
\end{thm}
\begin{proof}
Order the $\bar{v}$s and $\bar{w}$s so that $\bar{v}_1=\bar{w}_1,\dots,\bar{v}_t=\bar{w}_t$. Moving all terms to one side then gives
\[
(a_1-b_1)\bar{v}_1+\dots+(a_t-b_t)\bar{v}_t+a_{t+1}\bar{v}_{t+1}+\dots-(b_{t+1}\bar{w}_{t+1}+\dots)=\bar{0}.
\]
Since $X$ is assumed to be linearly independent, we learn that $a_{t+1}=\dots=a_{n}=0$ and $b_{t+1}=\dots=0$, and by our assumption on the non-zeroness of the coefficients, this forces $t=n=m$. Moreover, we learn that $a_i=b_i$ for all $i$.
\end{proof}
Thus knowing that $\bar{0}$ has a unique presentation as a linear combination ensures that everything in the span does so. We can use span more directly to test for linear independence.
\begin{prop}
A set $X$ is linearly independent if and only if for all $\bar{v}\in X$, $\bar{v}\not\in \langle X-\{\bar{v}\}\rangle$.
\end{prop}
\begin{proof}
We show the negative of this. $X$ is linearly dependent if there is a linear dependence relation
\[
a_1\bar{v}_1+\dots+a_n\bar{v}_n=\bar{0},
\]
with some coefficient (say $a_1$) not equal to zero. Solving for $\bar{v}_1$ then expresses $\bar{v}_1$ as an element of the span of the remaining vectors.
\end{proof}

Thus if a set is linearly independent, then we can add any vector not in the span of it to it and still have a linearly independent set. On the other hand, if we add in a vector in the span, then the set becomes linearly dependent.
\end{document} 