# Linear Algebra
2020 October
by: Emanuel Becerra Soto
----
## Topics
+ Subspaces
+ Bases
+ Matrices
+ Systems of Linear Equations
---
### Span
#### Definition
The set of all _linear combinations_ of a list of vectors in $V$ is called the _span_ of such list.
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### Span
#### Definition
$$
\text{span}\ (v_1, ..., v_m) = \{ a_1 v_1 + ... + a_m v_m \}
$$
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### Span
#### Intuition
1. The building of a _Subspace_ from a _List of Vectors_, almost by force.
2. To where in the space a basic list of directions can take me?
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### Span
#### Definition "Spans"
If the _span_ of a list of vectors equals $V$, we say that such list **spans** $V$.
----
### Span
#### Definition "Spans"
$$
\text{span}\ (v_1, v_2, ..., v_m) = V
$$
----
### Span
#### Exercises
+ **Span is a subspace.** (Hint: Check sufficient and necessary conditions for subspaces.)
+ **Span is the smallest containing subspace.** (Hint: Verify that Span is a subset of all other containing subspaces.)
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### Span
#### References
+ Axler, S. (2015). Linear algebra done right. springer. **Chapter 2**
---
### Basis
#### Definition
A basis of $V$ is a list of vectors in $V$, that are _linearly independent_ and _spans_, $V$.
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### Basis
#### Definition
$$
\text{basis} \big[l, V \big] :\Leftrightarrow \text{l.i} \big[ l \big] \wedge \text{spans} \big[ l, V \big]
$$
$$
l = (v_1, ..., v_m)
$$
----
### Basis
#### References
+ Axler, S. (2015). Linear algebra done right. springer. **Chapter 2**
---
### Dimension
#### Basis Length does not depend on basis
**Proposition.** Any two bases of a finite-dimensional vector space have the same length.
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### Dimension
#### Proof. Basis Length does not depend on basis
+ Lemma 2.23: Length of _l.i._ list $\leq$ length of spanning list
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### Dimension
#### Proof. Basis Length does not depend on basis
Let $B_1$ and $B_2$ be two bases of $V$.
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### Dimension
#### Proof. Basis Length does not depend on basis
By definition $l.i.\ [B_1] \equiv True$ and $spans\ [B_2] \equiv True$
So by Lemma 2.23 $\text{len}\ (B_1) \leq \text{len}\ (B_2)$.
----
### Dimension
#### Proof. Basis Length does not depend on basis
By a similar reason we have $\text{len}\ (B_2) \leq \text{len}\ (B_1)$.
----
### Dimension
#### Proof. Basis Length does not depend on basis
$\text{len}\ (B_1) \leq \text{len}\ (B_2)$
$\text{len}\ (B_2) \leq \text{len}\ (B_1)$
+ $\text{len}\ (B_1) = \text{len}\ (B_2)$
----
### Dimension
#### Definition
The **dimension** of a finite-dimensionl vector space $V$ is the length of any basis of $V$.
----
#### Dimension
#### Notation
The **dimension** of $V$ is denoted as:
+ $\text{dim}\ V$
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
In a finite-dimensional vector space, the length of every _linearly independent_ list of vectors is less than or equal to the length of every _spanning_ list of vector.
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### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
#### Strategy to prove it
Slowly, but surely, turn the _spanning list_ to the _l.i. list_.
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### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
Let's call the the _l.i. list_ $l_A$ and the _spanning list_ $l_B$.
$$
l_A = (u_1, ... , u_m)
$$
$$
l_B = (v_1, ... , v_n)
$$
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
We need to show that:
$$
m \leq n
$$
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
##### Step 1
Add an abitrary $u_i$ from $l_A$ to $l_B$. Let's say it was $u_1$.
$$
u_1, w1, ..., w_n, \text{length} = n + 1
$$
----
##### Step 1
$$
u_1, w1, ..., w_n, \text{length}\ [l_B] = n + 1
$$
+ The new list is _linearly dependant_!
+ Using the _Linear Dependence Lemma_ we can remove a vector from the new list.
+ Let's remove a $w_i$ and the new list now has a length of $n$.
+ And still _spans_ $V$.
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
##### Step j
$l_B$ from the step $j-1$ has size $n$.
Add an abitrary $u_j$ from $l_A$ to $l_B$.
$$
u_1, ..., u_j, w1, ..., w_n, \text{length}\ [l_B] = n + 1
$$
----
##### Step j
$l_B$ from the step $j-1$ has size $n$.
Add an abitrary $u_j$ from $l_A$ to $l_B$.
$$
u_1, ..., u_j, w1, ..., w_n, \text{length}\ [l_B] = n + 1
$$
+ The new list is _linearly dependant_!
+ And because all the previously added $u_j$'s are $l.i.$
+ We are able to find a $w_i$ that we can remove
+ While our list still spans $V$.
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
This process terminates when the last $u_m$ was added and thus the length of $l_B$ was at least the lenght of $l_A$.
$$
\text{len}\ l_A \leq l_B
$$
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
The _Linear Dependece Lemma_ guarantees that we can find a $w_j$ at each step.
----
### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
#### Examples
Show that the list (1,2,3), (4,5,6), (9,6,7), (-3,2,8) is not _l.i._ in $\mathbb{R}^3$.
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### Dimension
#### **Lemma 2.23:** Length of _l.i._ list $\leq$ length of spanning list
#### Examples
Show that the list (1,2,3), (4,5,6), (9,6,7), (-3,2,8) is not _l.i._ in $\mathbb{R}^3$.
As we know the list (1,0,0), (0,1,0), (0,0,1) spans $\mathbb{R}^3$, so no list of length larger than 3 is l.i. in $\mathbb{R}^3$
----
### Dimension
#### **Lemma 2.21:** Linear Dependence Lemma
#### Exersise
Suppose $v_1,...,v_m$ is a _l.d._ list in $V$. Then there exists a $j \in {1,2,...,m}$ such that the following holds:
1. $v_j \in \text{span}\ (v_1, v_2, ... ,v_{j-1})$
2. If $v_j$ is removed from the list, the remaining list span is equal to $\text{span}\ (v_1, v_2, ... ,v_m)$
_Hint:_ Find a vector that is a linear combination of the others.
----
### Dimension
#### References
+ Axler, S. (2015). Linear algebra done right. springer. **Chapter 2**
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