Lecture 10 - Finishing LI&LD, Bases

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Midterm is pushed back. Check canvas later on but yeah.

Finishing 2.A

Theorem

In a vector space $V$ with $v_{1}, . . ., v_{k}$ are LD list:

There is some $v_{j}$ such that $v_{j}$ is a linear combination of $v_{1}, . . ., v_{j - 1}$
$span (v_{1}, . . ., v_{k}) = span (v_{1}, . . ., v_{j - 1}, v_{j + 1}, . . ., v_{k})$ .

Note that this makes sense, since you can change the order of vectors (not the contents) such that the span is the same.

Linear Independence Theorem

The length of a LI list is always $\leq$ to the length of a spanning list in any vector space $V$ .

Proof
Suppose $u_{1}, . . ., u_{m}$ is a LI list and that $v_{1}, . . ., v_{n}$ is a spanning list for $V$ . We want to show that $m \leq n$ .

Consider $v_{1}, . . ., v_{n}$ . We can prepend $u_{1}$ to this list. This new list is dependent since the $v^{'} s$ span $V$ , so the $u_{1}$ can be written in terms of $v_{1}, . . ., v_{n}$ . So $u_{1}, v_{1}, . . ., v_{n}$ is a LD list. Since it is, then there's some vector in the list that can be written as the previous, and if we remove the vector is doesn't change the span. Thus, there's some $v_{i}$ that is a linear combination of previous vectors. We throw out $v_{i}$ and consider the new list.

Consider $u_{1}, v_{1}, . . ., v_{i - 1}, v_{i + 1}, . . ., v_{n}$ , whose span has been unchanged from before. Repeat this idea, and consider $u_{2}, u_{1}, . . ., v_{i - 1}, v_{i + 1}, . . ., v_{n}$ . It is a LD list, so one of the vectors is a combination of the previous. It has to be one of the $v$ 's again since the $u$ 's are LI. So there is again some $v_{j}$ that is a linear combination of the previous. We throw it out and repeat.

Note then that there must be at least as many $v$ 's as there are $u$ 's, so $m \leq n$ , and the proof is done.
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Note

Recall that $U \equiv$ subspace of $2 \times 2$ symmetric matrices:

[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]

So any list of length 4 or more vectors $u_{1}, . . ., u_{4} \in U$ is guaranteed to be LD.

Bases

Consider the diagram below:

Thus having both properties are interesting to us, so we make a definition

Basis

A list $v_{1}, . . ., v_{k}$ in a vector space $V$ is a basis of $V$ if the list is:

L.I.
A spanning list

Now suppose that $u_{1}, . . ., u_{m}$ is a basis for $V$ , and $v_{1}, . . ., v_{n}$ is also a basis for $V$ . Using the LI Theorem, then $m \leq n$ , but we can do it the other way so then $n \leq m$ , so $n = m$ . Therefore, all bases for a vector space are the same number, so we give it a name:

Dimension

The number of vectors in a basis of $V$ is called the dimension of $V$ .

So if I have a LI list that isn't spanning, then the number of vectors is too low. And if a spanning list is LD, then we need to remove some vectors:

Hence when adding new vectors to make a LI list still LI is an extension to basis, and likewise removing vectors from a spanning list to make it LI is a reduction to basis (trimming).

Reduction

Suppose $v_{1}, . . ., v_{n}$ is a spanning list of $V$ . If the list of $v_{1}, . . ., v_{n}$ is LI then we are done. Otherwise, then the list must be LD, so at least one of these $v_{i}$ can be made as a linear combination of the others. Thus, we can throw it out without changing the span:

v_{1}, . . ., v_{i - 1}, v_{i + 1}, . . ., v_{n}

Thus, remove it, and repeat this process until we are LI. This must eventually terminate as there's a finite number of vectors in the list.

Extension

Suppose $v_{1}, . . ., v_{m}$ is LI. If they span, great we are done. If not, add a vector outside of the span of the list, say $u_{i}$ . The new list must be LI, so we can just repeat. Since $m \leq n$ , in a finitely-dimensional vector space, we must terminate once we have $n$ vectors in our list.