06-240/Classnotes For Thursday, September 21: Difference between revisions

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==<center><u>'''Force Vectors'''</u></center>==
==<center><u>'''Force Vectors'''</u></center>==
#There is a special force vector called 0.
#<math>\mbox{There is a special force vector called 0.}</math>
#They can be added.
#<math>\mbox{They can be added.}</math>
#They can be multiplied by any scalar.
#<math>\mbox{They can be multiplied by any scalar.}</math>


====''Properties''==== (convention: x,y,z-vectors; a,b,c-scalars)
====''Properties''====

# <math> x+y=y+x \ </math>
<math>\mbox{(convention: }x,y,z \mbox{ are vectors; }a,b,c \mbox{ are scalars)}</math>
#<math> x+y=y+x \ </math>
#<math> x+(y+z)=(x+y)+z \ </math>
#<math> x+(y+z)=(x+y)+z \ </math>
#<math> x+0=x \ </math>
#<math> x+0=x \ </math>
#<math> \forall x\; \exists\ y \ s.t.\ x+y=0 \ </math>
#<math> \forall x\; \exists\ y \ \mbox{ s.t. }x+y=0</math>
#<math> 1\cdot x=x \ </math>
#<math> 1\cdot x=x \ </math>
#<math> a(bx)=(ab)x \ </math>
#<math> a(bx)=(ab)x \ </math>
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#<math> (a+b)x=ax+bx \ </math>
#<math> (a+b)x=ax+bx \ </math>


=====Definition=====
=====Definition===== Let F be a field "of scalars". A vector space over F is a set V (of "vectors") along with two operations:

Let F be a field "of scalars". A vector space over F is a set V, of "vectors", along with two operations

: <math> +: V \times V \to V </math>
: <math> +: V \times V \to V </math>
: <math> \cdot: F \times V \to V </math>, so that
: <math> \cdot: F \times V \to V \mbox{, so that:}</math>
#<math> \forall x,y \in V\ x+y=y+x </math>
#<math> \forall x,y \in V\ x+y=y+x </math>
#<math> \forall x,y \in V\ x+(y+z)=(x+y)+z </math>
#<math> \forall x,y \in V\ x+(y+z)=(x+y)+z </math>
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<math> 0_{F^n}=(0,\ldots,0) </math> <br/>
<math> 0_{F^n}=(0,\ldots,0) </math> <br/>
<math> a\in F\ ax=(aa_1,aa_2,\ldots,aa_n) </math> <br/>
<math> a\in F\ ax=(aa_1,aa_2,\ldots,aa_n) </math> <br/>
<math> In \ \mathbb{Q}^3 \ \left( \frac{3}{2},-2,7\right)+\left( \frac{-3}{2}, \frac{1}{3},240\right)=\left(0, \frac{-5}{3},247\right) </math> <br/>
<math> \mbox{In } \mathbb{Q}^3 \ \left( \frac{3}{2},-2,7\right)+\left( \frac{-3}{2}, \frac{1}{3},240\right)=\left(0, \frac{-5}{3},247\right) </math> <br/>
<math> 7\left( \frac{1}{5},\frac{1}{7},\frac{1}{9}\right)=\left( \frac{7}{5},1,\frac{7}{9}\right) </math> <br/>
<math> 7\left( \frac{1}{5},\frac{1}{7},\frac{1}{9}\right)=\left( \frac{7}{5},1,\frac{7}{9}\right) </math> <br/>
'''Ex.2.'''
'''Ex.2.'''
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& \vdots \\ a_{m1} & \cdots & a_{mn}\end{pmatrix}: a_{ij} \in F \right\rbrace </math> <br/>
& \vdots \\ a_{m1} & \cdots & a_{mn}\end{pmatrix}: a_{ij} \in F \right\rbrace </math> <br/>
<math> M_{3\times 2}( \mathbb{R})\ni \begin{pmatrix} 7 & -7 \\ \pi & \mathit{e} \\ -5 & 2 \end{pmatrix} </math> <br/>
<math> M_{3\times 2}( \mathbb{R})\ni \begin{pmatrix} 7 & -7 \\ \pi & \mathit{e} \\ -5 & 2 \end{pmatrix} </math> <br/>
Addition by adding entry by entry:
<math>\mbox{Addition by adding entry by entry:}</math>


<math> M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}+\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \end{pmatrix}</math> <br/>
<math> M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}+\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \end{pmatrix}</math> <br/>


Multiplication by multiplying scalar c to all entries by M.
<math>\mbox{Multiplication by multiplying scalar c to all entries by M.}</math>


<math> c\cdot M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} c\cdot a_{11} & c\cdot a_{12} \\ c\cdot a_{21} & c\cdot a_{22} \end{pmatrix}</math> <br/> <br/>
<math> c\cdot M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} c\cdot a_{11} & c\cdot a_{12} \\ c\cdot a_{21} & c\cdot a_{22} \end{pmatrix}</math> <br/> <br/>


Zero matrix has all entries = 0:
<math>\mbox{Zero matrix has all entries = 0:}</math>


<math> 0_{M_{m\times n}}=\begin{pmatrix} 0 & \cdots & 0 \\ \vdots &
<math> 0_{M_{m\times n}}=\begin{pmatrix} 0 & \cdots & 0 \\ \vdots &
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<math> \mathbb{C}</math> form a vector space over <math> \mathbb{R}</math>. <br/>
<math> \mathbb{C}</math> form a vector space over <math> \mathbb{R}</math>. <br/>
'''Ex.4.'''
'''Ex.4.'''
F is a vector space over itself. <br/>
<math>\mbox{F is a vector space over itself.}</math> <br/>
'''Ex.5.'''
'''Ex.5.'''
<math> \mathbb{R}</math> is a vector space over <math> \mathbb{Q}</math>. <br/>
<math> \mathbb{R}</math> is a vector space over <math> \mathbb{Q}</math>. <br/>
'''Ex.6.'''
'''Ex.6.'''
Let S be a set. Let <br/>
<math>\mbox{Let S be a set. Let}</math> <br/>
<math> \mathcal{F}(S,\mathbb{R})=\big\{f:S\to \mathbb{R} \big\} </math> <br/>
<math> \mathcal{F}(S,\mathbb{R})=\big\{f:S\to \mathbb{R} \big\} </math> <br/>
<math> f,g \in \mathcal{F}(S,\mathbb{R}) </math> <br/>
<math> f,g \in \mathcal{F}(S,\mathbb{R}) </math> <br/>

Revision as of 05:31, 27 September 2006

A force has a direction & a magnitude.

Force Vectors

Properties

Definition

Let F be a field "of scalars". A vector space over F is a set V, of "vectors", along with two operations


9.

Examples

Ex.1.







Ex.2.





Ex.3. form a vector space over .
Ex.4.
Ex.5. is a vector space over .
Ex.6.