# 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. ---- ### Span #### Definition $$ \text{span}\ (v_1, ..., v_m) = \{ a_1 v_1 + ... + a_m v_m \} $$ ---- ### 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? ---- ### 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.) ---- ### 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_, $V$. ---- ### 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. ---- ### Dimension #### Proof. Basis Length does not depend on basis + Lemma 2.23: Length of _l.i._ list $\leq$ length of spanning list ---- ### Dimension #### Proof. Basis Length does not depend on basis Let $B_1$ and $B_2$ be two bases of $V$. ---- ### 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. ---- ### 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_. ---- ### 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 $\mathbb{R}^3$. ---- ### 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**