# 06-240/Classnotes For Tuesday October 3

Definition. ${\displaystyle v\in V}$is a linear combination of elements in ${\displaystyle S\subset V}$ if ${\displaystyle \exists u_{1},\ldots ,u_{n}\in S}$ and ${\displaystyle a_{1},\dots ,a_{n}\in F}$ such that ${\displaystyle V=\sum a_{i}u_{i}}$

Example. In ${\displaystyle P_{3}(\mathbb {R} )}$, ${\displaystyle v_{1}=2x^{3}-2x^{2}+12-6}$ is a linear combination of: ${\displaystyle u_{1}=x^{3}-2x^{2}-5x-3}$ and ${\displaystyle u_{2}=3x^{3}-5x^{2}-4x-9}$ but ${\displaystyle v_{2}=3x^{3}-2x^{2}+7x+8}$ is not.

Why? ${\displaystyle v_{1}=2x^{3}-2x^{2}+12-6=a_{1}u_{1}+a_{2}u_{2}}$ ${\displaystyle =a_{1}(x^{3}-2x^{2}-5x-3)+a_{2}(3x^{3}-5x^{2}-4x-9)}$

${\displaystyle v_{1}^{}=-4u_{1}+2u_{2}}$

${\displaystyle {\mbox{Definition}}{}_{}^{}}$

${\displaystyle {\mbox{We say that a subset }}S\subset V{\mbox{ generates or spans }}V}$

${\displaystyle {\mbox{ if span }}S=\lbrace {\mbox{ all linear combinations of elements in }}S\rbrace =V{}_{}^{}}$

${\displaystyle {\mbox{Examples}}{}_{}^{}}$

${\displaystyle V=M_{2\times 2}(\mathbb {R} )}$

${\displaystyle M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&1\\0&0\end{pmatrix}},M_{3}={\begin{pmatrix}0&0\\1&0\end{pmatrix}},M_{4}{\begin{pmatrix}0&0\\0&1\end{pmatrix}}}$

${\displaystyle N_{1}={\begin{pmatrix}0&1\\1&1\end{pmatrix}},N_{2}={\begin{pmatrix}1&0\\1&1\end{pmatrix}},N_{3}={\begin{pmatrix}1&1\\0&1\end{pmatrix}},N_{4}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}}$

${\displaystyle {\mbox{Claims}}{}_{}^{}}$

1. ${\displaystyle \lbrace M_{1}^{},M_{2},M_{3},M_{4}\rbrace {\mbox{ generates }}V}$
2. ${\displaystyle \lbrace N_{1}^{},N_{2},N_{3},N_{4}\rbrace {\mbox{ generates }}V}$
3. ${\displaystyle \lbrace M_{1}^{},M_{2},M_{3}\rbrace {\mbox{ does not generate }}V}$
4. ${\displaystyle \lbrace N_{1}^{},N_{2},N_{3}\rbrace {\mbox{ does not generate }}V}$

${\displaystyle {\mbox{Proof of 1}}{}_{}^{}}$

${\displaystyle {\mbox{Given any }}B={\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}{\mbox{ need to find }}a_{1},a_{2},a_{3},a_{4}{\mbox{ such that,}}}$

${\displaystyle {\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}=B=a_{1}M_{1}+a_{2}M_{2}+a_{3}M_{3}+a_{4}M_{4}={\begin{pmatrix}a_{1}&0\\0&0\end{pmatrix}}+{\begin{pmatrix}0&a_{2}\\0&0\end{pmatrix}}+{\begin{pmatrix}0&0\\a_{3}&0\end{pmatrix}}+{\begin{pmatrix}0&0\\0&a_{4}\end{pmatrix}}}$

${\displaystyle ={\begin{pmatrix}a_{1}^{}&a_{2}\\a_{3}&a_{4}\end{pmatrix}}\Leftrightarrow {\begin{cases}b_{11}=a_{1}\\b_{12}=a_{2}\\b_{21}=a_{3}\\b_{22}=a_{4}\end{cases}}}$ ${\displaystyle {\mbox{A system of 4 equations with 4 unknowns}}{}_{}^{}}$
${\displaystyle {\mbox{Proof of 2}}{}_{}^{}}$

${\displaystyle {\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}=B=a_{1}N_{1}+a_{2}N_{2}+a_{3}N_{3}+a_{4}N_{4}={\begin{pmatrix}0&a_{1}\\a_{1}&a_{1}\end{pmatrix}}+{\begin{pmatrix}a_{2}&0\\a_{2}&a_{2}\end{pmatrix}}+{\begin{pmatrix}a_{3}&a_{3}\\0&a_{3}\end{pmatrix}}+{\begin{pmatrix}a_{4}&a_{4}\\a_{4}&0\end{pmatrix}}}$

${\displaystyle ={\begin{pmatrix}a_{2}^{}+a_{3}+a_{4}&a_{1}+a_{3}+a_{4}\\a_{1}+a_{2}+a_{4}&a_{1}+a_{2}+a_{3}\end{pmatrix}}\Leftrightarrow {\begin{cases}b_{11}=a_{2}+a_{3}+a_{4}\\b_{12}=a_{1}+a_{3}+a_{4}\\b_{21}=a_{1}+a_{2}+a_{4}\\b_{22}=a_{1}+a_{2}+a_{3}\end{cases}}}$

${\displaystyle {\mbox{Trick}}{}_{}^{}}$

${\displaystyle M_{1}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{1}}$ ${\displaystyle M_{2}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{2}}$ ${\displaystyle M_{3}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{3}}$ ${\displaystyle M_{4}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{4}}$

${\displaystyle B=b_{11}^{}M_{1}+b_{12}M_{3}+b_{21}M_{3}+b_{22}M_{4}}$ ${\displaystyle =b_{11}\left({\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{1}\right)+\ldots }$

${\displaystyle ={\mbox{ a linear combination of }}N_{1}^{},N_{2},N_{3},N_{4}}$

${\displaystyle {\mbox{Proof of 3}}{}_{}^{}}$

${\displaystyle {\mbox{Indeed in }}a_{1}^{}M_{1}+a_{2}M_{2}+a_{3}M_{3}={\begin{pmatrix}a_{1}&a_{2}\\a_{3}&0\end{pmatrix}}{\mbox{ lower right corner is always }}0}$

${\displaystyle {\mbox{for example }}{\begin{pmatrix}240&157\\e&\pi \end{pmatrix}}{\mbox{ not in span.}}}$

${\displaystyle {\mbox{Proof of 4}}{}_{}^{}}$

${\displaystyle a_{1}^{}N_{1}+a_{2}N_{2}+a_{3}N_{3}={\begin{pmatrix}a_{2}+a_{3}&a_{1}+a_{3}\\a_{1}+a_{2}&a_{1}+a_{2}+a_{3}\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}240&157\\e&\pi \end{pmatrix}}{\mbox{ is equal? }}{\begin{cases}240=a_{2}+a_{3}\\157=a_{1}+a_{3}\\e=a_{1}+a_{2}\\\pi =a_{1}+a_{2}+a_{3}\end{cases}}\Rightarrow {\mbox{No solution}}}$

${\displaystyle {\mbox{Motivation}}{}_{}^{}}$

${\displaystyle S\subset V{\mbox{ is linearly dependent if it is wasteful,}}}$ ${\displaystyle {\mbox{i.e. if }}\exists v\in V{\mbox{ such that }}\exists a_{1}^{}\ldots a_{n}\in F{\mbox{ and }}u_{1}^{}\ldots u_{2}\in S}$ ${\displaystyle {\mbox{ and }}\exists b_{1}^{}\ldots b_{m}\in F{\mbox{ and }}w_{1}\ldots w_{m}\in S}$

${\displaystyle {\mbox{so that }}\sum _{i=1}^{n}a_{i}u_{i}=v=\sum _{i=1}^{m}b_{i}w_{i}}$

${\displaystyle \sum a_{i}u_{i}-\sum b_{i}w_{i}=0}$

${\displaystyle {\mbox{can be represented as }}\sum c_{i}z_{i}=0}$

${\displaystyle {\mbox{Definition}}{}_{}^{}}$

${\displaystyle S\subset V{\mbox{ is called linearly dependent if you can find }}}$ ${\displaystyle z_{1}^{}\ldots z_{n}\in S{\mbox{ different from each other and }}c_{1}^{}\ldots c_{n}\in F{\mbox{ so that not all of which are }}0,}$ ${\displaystyle {\mbox{so that }}\sum c_{i}z_{i}=0{\mbox{ otherwise, }}S{\mbox{ is called linearly independent}}}$

${\displaystyle {\mbox{Example 1}}{}_{}^{}}$

${\displaystyle {\mbox{In }}\mathbb {R} ,S=\lbrace {\begin{pmatrix}1&2&3\end{pmatrix}},{\begin{pmatrix}4&5&6\end{pmatrix}},{\begin{pmatrix}7&8&9\end{pmatrix}}\rbrace {\mbox{ is linearly dependent}}}$

${\displaystyle 1\cdot {\begin{pmatrix}1&2&3\end{pmatrix}}-2\cdot {\begin{pmatrix}4&5&6\end{pmatrix}}+1\cdot {\begin{pmatrix}7&8&9\end{pmatrix}}=0}$

${\displaystyle {\mbox{Example 2}}{}_{}^{}}$

${\displaystyle \mathbb {R} ^{n},e_{i}={\begin{pmatrix}0\\\vdots \\1\\\vdots \\0\end{pmatrix}}i^{th}{\mbox{ row}}}$

${\displaystyle S=\lbrace e_{1}^{},\ldots ,e_{n}\rbrace }$

${\displaystyle {\mbox{Claim }}S{\mbox{ is linearly independent}}{}_{}^{}}$

${\displaystyle {\begin{pmatrix}0\\\vdots \\0\end{pmatrix}}=0=\sum _{i=1}^{n}a_{i}e_{i}={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}\Rightarrow {\begin{matrix}a_{1}=0\\a_{2}=0\\\vdots \\a_{n}=0\end{matrix}}}$

${\displaystyle {\mbox{not not all }}a_{i}^{}{\mbox{ are }}0\Rightarrow {\mbox{ not linearly dependent.}}}$

${\displaystyle {\mbox{Claim }}S\subset V{\mbox{ is linearly independent iff whenever }}\sum a_{i}u_{i}=0{\mbox{ and distinct }}u_{i}\in S{\mbox{ then }}\forall i\quad a_{i}=0}$

${\displaystyle {\mbox{Comments}}{}_{}^{}}$

1. ${\displaystyle \emptyset \subset V{\mbox{ is linearly independent}}}$
2. ${\displaystyle {\mbox{Suppose }}u\in V,\quad \lbrace u\rbrace {\mbox{ the singleton set is linearly independent iff }}u_{}^{}\neq 0}$

${\displaystyle \lbrace 0\rbrace {\mbox{ is linearly dependent. example }}7\cdot 0=0}$

${\displaystyle {\mbox{if }}u\neq 0{\mbox{ assume }}a\cdot u=0{\mbox{, and }}a\neq 0\Rightarrow a_{}^{-1}au=0\Rightarrow u=0{\mbox{ contradiction results, so no such }}a{\mbox{ exists.}}}$ ${\displaystyle {\mbox{ So}}{}_{}^{}\lbrace u\rbrace {\mbox{is not linearly dependent, hence it is linearly independent.}}}$