We'll cover:

• Span
• Linear Independence
• Bases
• Dimension

2.A: Span and Linear Independence

We'll sometimes need to consider lists of vectors, which we usually write without the surrounding parenthesis, ie: $(4,1,6),(9,5,7)$ is a list of 2 vectors containing vectors in ${\mathbb{R}}^3$.

Linear Combinations and Span

So for instance, $(17,-4,2)$ is a linear combination of $(2,1,-3),(1,-2,4)$ since:

If instead a linear combination doesn't exist, then the related system of equations has no solutions.

So then, using our example above, $(17,-4,2)\in \text{span}((2,1,-3),(1,-2,4))$. Sometimes linear span is used to mean span.

Proof
Suppose $v_1,...,v_m$ is a list of vectors in $V$.

First, we show that $\text{span}(v_1,...,v_m)$ is a subspace of $V$. Clearly $\overrightarrow{0}\in \text{span}$ since:

Also the span is closed under addition since:

And also closed under scalar multiplication since:

Thus $\text{span}(v_1,...,v_m)$ is a subspace of $V$.

Each $v_i$ is a linear combination from $v_1,..,v_m$ by just adding the $v_i$ from the set itself. Conversely, due to closure under vector addition and scalar multiplication, every subspace of $V$ containing each $v_j$ contains $\text{span}(v_1,...,v_m)$. So then the span is the smallest subspace of $V$ containing all vectors $v_1,...,v_m$.
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If $\text{span}(v_1,...,v_m)$ equals $V$ then we say that $v_1,...,v_m$ spans $V$.

So clearly ${\mathbb{F}}^n$ is finite dimensional.

You can show that $\mathcal{P}(\mathbb{F})$ is a vector space, and thus is a subspace of ${\mathbb{F}}^{\mathbb{F}}$.

You can verify ${\mathcal{P}}_m(\mathbb{F})$ is a finite-dimensional vector space for each non-negative integer $m$.

We can, for instance, show that $\mathcal{P}(\mathbb{F})$ is infinite-dimensional via the following. Consider any list of elements of $\mathcal{P}(\mathbb{F})$. Let $m$ denote the highest degree of polynomials in this list. Then every polynomial in the span of this list has degree at most $m$. Thus, $z^{m+1}$ is not in the span of our list. Hence, no list spans $\mathcal{P}(\mathbb{F})$, so it is infinite-dimensional.

Linear Independence

Suppose $v_1,...,v_m\in V$ and $v\in \text{span}(v_1,...,v_m)$. By the definition of span, there exist $a_1,...,a_m\in \mathbb{F}$ such that:

Consider the question of whether the choice of scalars in the equation above is unique. Suppose $c_1,...,c_m$ is another set of scalars such that:

Subtracting the two equations:

Thus we have written 0 as a linear combination of $v_1,...,v_m$. If the only way to do this is via is via $0=0v_1+\cdots +0v_m$ then each $a_i$ equals $c_i$, and thus the choice of scalars is unique. We give this idea a name:

A list of vectors in $V$ is called linearly dependent if it is not linearly independent. In other words, a list $v_1,...,v_m$ of vectors in $V$ is linearly dependent if there exist $a_1,...,a_m\in \mathbb{F}$, not all 0, such that $a_1{v}_1+\cdots +a_m{v}_m=0$.

The lemma below will often be useful:

Proof
Because the list $v_1,...,v_m$ is linearly dependent, then there exist numbers $a_1,...,a_m\in \mathbb{F}$, not all 0, such that:

Let $j$ be the largest element of $\{1,...,m\}$ such that $a_j\ne 0$. Then:

Proving the first part. To prove the second, suppose $u\in \text{span}(v_1,...,v_m)$, so then there exist numbers $c_1,...,c_m\in \mathbb{F}$ such that:

We can replace $v_j$ with the right side of the equation defining $v_j$, showing that the $u$ is in the span of the list obtained by removing the $j$-th term from the original list. Thus, the second condition holds.
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Note that the proof doesn't call into question the cases where some $v_i=0$, but those special cases can definitely still be remedied. We won't call to attention cases with:

• Empty lists
• Lists of length 1
• The subspace $\{0\}$
• ...

The proof is at Year3/Winter2024/MATH306-LinearAlgebraII/2015_Book_LinearAlgebraDoneRight.pdf#pages=35. We can use this idea to show that a list is linearly dependent since, given some list of linearly independent vectors that span $V$, the list in question has "too many" vectors in comparison.

Our intuition suggests that every subspace of a finite-dimensional vector space should also be finite-dimensional.

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. We'll prove that $U$ is finite-dimensional.

• Step 1: If $U=\{0\}$ then $U$ is finite-dimensional and we are done. If $U\ne \{0\}$ then choose a nonzero vector $v_1\in U$.
• Step $j$: If $U=\text{span}(v_1,...,v_{j-1})$ then $U$ is finite-dimensional and we are done. If $U\ne \text{span}(v_1,...,v_{j-1})$ then choose a vector $v_j\in U$ such that: $$
  v_j \notin \text{span}(v_1, ..., v_{j-1})

v = a_1v_1 + \dots + a_nv_n

$$where\$a_1,...,a_n\in \mathbb{F}\$.$$

Proof
First, suppose $v_1,...,v_n$ is a basis of $V$. Let $v\in V$. Because $v_1,...,v_n$ spans $V$, there exist $a_1,...,a_n\in \mathbb{F}$ such that $v=a_1{v}_1+\cdots +a_n{v}_n$ holds. To show the representation is unique, suppose $c_1,...,c_n$ are different scalars such that:

Subtracting the last equation from the first gives:

But since $v_1,...,v_n$ is LI, then all $a_i-c_i=0\Rightarrow a_i=c_i$ so then we have achieved uniqueness.

Now for the other direction. Suppose every $v\in V$ can be written uniquely in the form:

This clearly implies that $v_1,...,v_n$ spans $V$. To show that $v_1,...,v_m$ is LI, suppose that $a_1,...,a_n\in \mathbb{F}$ such that:

by applying $v=\overrightarrow{0}$. Since the representation must be unique (as a given) then we must have it such that all $a_i=0$ so then $v_1,...,v_m$ is LI.
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Clearly some spanning lists have too many vectors for them to be a basis (ie: LI). As an example, in the space ${\mathbb{F}}^2$ we may start with $(1,2),(3,6),(4,7),(5,9)$. We can use the following proof to remove vectors to remain with $(1,2),(4,7)$ as a valid basis.

Proof
Suppose $v_1,...,v_n$ spans $V$. We'll want to remove some of the vectors from $v_1,...,v_n$ so that the remaining vectors form a basis of $V$. We do this via the following process.

Start with $B=\{v_1,...,v_n\}$. If $v_1=\overrightarrow{0}$, delete it from $B$. If $v_1\ne \overrightarrow{0}$, then leave $B$ unchanged.
Repeat, where if $v_j$ is in $\text{span}(v_1,...,v_{j-1})$ then delete $v_j$ from $B$, otherwise leave $B$ unchanged.

After step $n$, we get a list $B$ which spans $V$ since our original list spanned $V$ and we only removed vectors that were already in the span of previous vectors. The process ensures that no vector in $B$ is in the span of the previous ones, so then $B$ is LI by the Chapter 2 - Finite-Dimensional Vector Spaces#^3867cf Lemma. So $B$ is a basis of $V$.
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This implies the following:

Proof
By definition, a finite-dimensional vector space has a spanning list. The previous proof shows how we can reduce such a list to a basis.
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Going the other way, we can try to add vectors to a LI list that's too small to ensure that it spans.

Proof
Suppose $u_1,...,u_m$ is LI in a finite-dimensional vector space $V$. Let $w_1,...,w_n$ be a basis of $V$. Thus the list:

spans $V$. Applying the procedure of finding our basis from Chapter 2 - Finite-Dimensional Vector Spaces#^e91abd, we can reduce this list to a basis of $V$ to produce a basis consisting of the vectors $u_1,...,u_m$, where no $u_i$ can get deleted as $u_1,...,u_m$ is LI, and some of the $w$'s.
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Proof
Because $V$ is finite-dimensional, so is $U$ via Chapter 2 - Finite-Dimensional Vector Spaces#^4a3f0a. Thus, there is a basis $u_1,...,u_m$ of $U$ via Chapter 2 - Finite-Dimensional Vector Spaces#^7eaea4. Of course, $u_1,...,u_m$ is a linearly independent list of vectors in $V$. Hence, this list can be extended to a basis $u_1,...,u_m,w_1,...,w_n$ of $V$ via Chapter 2 - Finite-Dimensional Vector Spaces#^aaf384. Let $W=\text{span}(w_1,...,w_n)$. To prove that $V=U\oplus W$, by Chapter 1 - Vector Spaces#^f3385e, then we must prove $U\cap W=\{\overrightarrow{0}\}$, and that $V=U+W$.

To prove $V=U+W$, suppose $v\in V$ is arbitrary. Then, because the list $u_1,...,u_m,w_1,...,w_n$ spans $V$ (as it's a basis), there exist $a_1,...,a_m,b_1,...,b_n\in \mathbb{F}$ such that:

In other words, we have $v=u+w$, where $u\in U$ and $w\in W$. Thus, $v\in U+W$, so then since $v$ was arbitrary it follows that $V=U+W$.

Now to show that $U\cap W=\{\overrightarrow{0}\}$. Suppose $v\in U\cap W$. Then there exist scalars $a_1,...,a_m,b_1,...,b_n\in \mathbb{F}$ such that:

as $v\in U$ and $v\in W$. Thus:

Because $u_1,...,u_m,w_1,...,w_n$ is LI (as it's a basis) then all $a_i,b_i=0$, so then $v=0$ completing the proof.
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2.C: Dimension

We know that something like ${\mathbb{F}}^n$ should be $n$ dimensional, but why? It comes from the fact that:

Proof
Suppose $V$ is finite-dimensional. Let $B_1,B_2$ be two bases of $V$. Both $B_1,B_2$ are LI and span $V$. But by Chapter 2 - Finite-Dimensional Vector Spaces#^ac1e1f then the length of $B_1$ is $\le$ that of $B_2$. Likewise, the length of $B_2$ is $\le$ to that of $B_1$, so then their lengths equal.
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Thus:

For instance, $\text{dim}({\mathbb{F}}^n)=n$. Further, ${\mathcal{P}}_m(\mathbb{F})=m+1$.

Proof
Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Think of a basis of $U$ as a LI list in $V$, and think of a basis of $V$ as a spanning list in $V$. Use Chapter 2 - Finite-Dimensional Vector Spaces#^ac1e1f to conclude that $\text{dim}(U)\le \text{dim}(V)$.
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Thus, we only need to check span or LI to get a basis if we know that the number of vectors is the dimension of the space we are working in.

Proof
Suppose $\text{dim}(V)=n$ and $v_1,...,v_n$ is a LI list in $V$. The list can be extended to a basis via Chapter 2 - Finite-Dimensional Vector Spaces#^aaf384, but in this case since every basis has a length $n$ then we do just the trivial extension. Thus, no elements are adjoined, so then $v_1,...,v_n$ is a basis of $V$.
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We can do the same for a spanning list:

Proof
Suppose $\text{dim}(V)=n$ and $v_1,...,v_n$ is a spanning list for $V$. The list can be reduced to a basis via Chapter 2 - Finite-Dimensional Vector Spaces#^e91abd, so then the reduction is trivial and no vectors are pruned. Thus, $v_1,...,v_n$ is a basis already for $V$.
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Proof
Let $u_1,...,u_m$ be a basis for $U_1\cap U_2$; thus $\text{dim}(U_1\cap U_2)=m$. Because $u_1,...,u_m$ is a basis of $U_1\cap U_2$, then it is LI in $U_1$, so the list can be extended to a basis $u_1,...,u_m,v_1,...,v_j$ of $U_1$ by Chapter 2 - Finite-Dimensional Vector Spaces#^aaf384. Thus, $\text{dim}(U_1)=m+j$. We can also extend $u_1,...,u_m$ to a basis $u_1,...,u_k,w_1,...,w_k$ of $U_2$, so then $\text{dim}(U_2)=m+k$.

We'll show that:

is a basis of $U_1+U_2$. This will complete the proof, as we'll have:

Clearly $u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$ spans $U_1$ and $U_2$ and hence equals $U_1+U_2$. To show that the list is a basis of $U_1+U_2$, we need to only show that it is LI (as we have the right number of basis vectors). So suppose:

where all $a_i,b_i,c_i$ are scalars. We'll need to prove all are equal to 0.

The equation above can be rewritten as:

so clearly then $c_1{w}_1+\cdots +c_k{w}_k\in U_1$. All the $w$'s are in $U_2$ so this implies that $c_1{w}_1+\cdots +c_k{w}_k\in U_1\cap U_2$. Because $u_1,...,u_m$ is a basis of $U_1\cap U_2$, we can write:

for some choice of scalars $d_1,...,d_m$. But $u_1,...,u_m,w_1,...,w_k$ is LI, so the last equation implies that all the $c$'s (and consequently all the $d$'s) equal 0. Thus our original equation becomes:

and since $u_1,...,u_m,v_1,...,v_j$ is LI, this equation implies that all the $a$'s and $b$'s are 0. Thus, all $a$'s, $b$'s, and $c$'s are 0 as desired.
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