06-240/Classnotes For Tuesday October 3

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Definition. v\in Vis a linear combination of elements in S\subset V if \exists u_1,\ldots,u_n\in S and a_1,\dots,a_n \in F such that V=\sum a_i u_i

Example. In P_3(\mathbb{R}), v_1=2x^3-2x^2+12-6 is a linear combination of: u_1=x^3-2x^2-5x-3 and u_2=3x^3-5x^2-4x-9 but v_2=3x^3-2x^2+7x+8 is not.

Why? v_1=2x^3-2x^2+12-6=a_1u_1+a_2u_2 =a_1(x^3-2x^2-5x-3)+a_2(3x^3-5x^2-4x-9)

v_1^{}=-4u_1+2u_2

\mbox{Definition}{}_{}^{}

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

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

\mbox{Examples}{}_{}^{}

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

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}

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}

\mbox{Claims}{}_{}^{}

  1. \lbrace M_1^{},M_2,M_3,M_4\rbrace\mbox{ generates }V
  2. \lbrace N_1^{},N_2,N_3,N_4\rbrace\mbox{ generates }V
  3. \lbrace M_1^{},M_2,M_3\rbrace\mbox{ does not generate }V
  4. \lbrace N_1^{},N_2,N_3\rbrace\mbox{ does not generate }V


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

\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,}

\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}=B=a_1M_1+a_2M_2+a_3M_3+a_4M_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}

=\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} \mbox{A system of 4 equations with 4 unknowns}{}_{}^{}
\mbox{Proof of 2}{}_{}^{}

\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}
=B=a_1N_1+a_2N_2+a_3N_3+a_4N_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}

=\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}

\mbox{Trick}{}_{}^{}

M_1=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_1 M_2=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_2 M_3=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_3 M_4=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_4

B=b_{11}^{}M_1+b_{12}M_3+b_{21}M_3+b_{22}M_4 =b_{11}\left(\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_1\right)+\ldots

=\mbox{ a linear combination of }N_1^{},N_2,N_3,N_4


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

\mbox{Indeed in }a_1^{}M_1+a_2M_2+a_3M_3=
\begin{pmatrix}a_1&a_2\\a_3&0\end{pmatrix}\mbox{ lower right corner is always } 0

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


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

a_1^{}N_1+a_2N_2+a_3N_3=\begin{pmatrix}a_2+a_3&a_1+a_3\\a_1+a_2&a_1+a_2+a_3\end{pmatrix}

\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}


\mbox{Motivation}{}_{}^{}

S\subset V\mbox{ is linearly dependent if it is wasteful,} \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 \mbox{ and }\exists b_1^{}\ldots b_m\in F \mbox{ and }w_1\ldots w_m\in S

\mbox{so that }\sum_{i=1}^na_iu_i=v=\sum_{i=1}^mb_iw_i

\sum a_iu_i-\sum b_iw_i=0

\mbox{can be represented as }\sum c_iz_i=0


\mbox{Definition}{}_{}^{}

S\subset V\mbox{ is called linearly dependent if you can find } 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, \mbox{so that }\sum c_iz_i=0 
\mbox{ otherwise, }S\mbox{ is called linearly independent}


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

\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}

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


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

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

S=\lbrace e_1^{},\ldots,e_n\rbrace

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

\begin{pmatrix}0\\\vdots\\0\end{pmatrix}=0
=\sum_{i=1}^na_ie_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}

\mbox{not not all }a_i^{}\mbox{ are }0\Rightarrow \mbox{ not linearly dependent.}

\mbox{Claim }S\subset V\mbox{ is linearly independent iff whenever }\sum a_iu_i=0
\mbox{ and distinct }u_i\in S\mbox{ then }\forall i\quad a_i=0


\mbox{Comments}{}_{}^{}

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

\lbrace0\rbrace\mbox{ is linearly dependent. example }7\cdot0=0

\mbox{if }u\neq0\mbox{ assume }a\cdot u=0\mbox{, and }a\neq0\Rightarrow a_{}^{-1}au=0
\Rightarrow u=0\mbox{ contradiction results, so no such }a\mbox{ exists.} \mbox{ So}{}_{}^{}\lbrace u\rbrace\mbox{is not linearly dependent, hence it is linearly independent.}