User:Eml5526.s11.team7.jin/HW2

=Problem 2.3=

Given
Use $$\left\{ \right\} $$as basis, let :{| style="width:100%" border="0" $$ \displaystyle
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 * style="width:95%" |

{\mathbf{w}} = \sum\limits_i {{\mathbf{a}}_i}

$$     (Eq 3.1) Then
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{array}{ccccccccccccccc} {{\mathbf{w}} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0}&{\forall {\mathbf{w}} \in {\mathbb{R}^n}} \end{array}

$$     (Eq 3.2) is equal to
 * 
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 * {| style="width:100%" border="0"

$$ \displaystyle
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 * style="width:95%" |

\begin{array}{ccccccccccccccc} {{\mathbf{w}} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0}&{\forall \left\{ {{\beta _1}, \ldots ,{\beta _n}} \right\} \in {\mathbb{R}^n}} \end{array}

$$     (Eq 3.3) Such that $${\mathbf{w}} = \sum\limits_i $$
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Find
Show Equation 3 is equivalent to
 * {| style="width:100%" border="0"

$$ \displaystyle
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 * style="width:95%" |

\begin{array}{ccccccccccccccc} {{{\mathbf{a}}_i} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0}&{\forall i = 1, \ldots ,n} \end{array}

$$     (Eq 3.4)
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Solution
Since$$ \left\{ {{\beta _1}, \ldots ,{\beta _n}} \right\} $$are arbitrary, let
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$$ \displaystyle
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{\beta _1} = 1,{\beta _2} = {\beta _3} = \cdots  = {\beta _n} = 0

$$     (Eq 3.5) Then
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 * {| style="width:100%" border="0"

$$ \displaystyle
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 * style="width:95%" |

{\mathbf{w}} = {{\mathbf{a}}_1}

$$     (Eq 3.6) Equation 3 can be written as
 * 
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 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{\mathbf{a}}_1} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0

$$     (Eq 3.7) Similarly, let
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$$ \displaystyle
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{\beta _1} = {\beta _2} = \cdots  = {\beta _{n - 1}} = 0,{\beta _n} = 1

$$     (Eq 3.8) Then
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$$ \displaystyle
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 * style="width:95%" |

{\mathbf{w}} = {{\mathbf{a}}_n}

$$     (Eq 3.9) Equation 3 can be written as
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$$ \displaystyle
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 * style="width:95%" |

{{\mathbf{a}}_n} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0

$$     (Eq 3.10) Thus, Equation 3 is identical to Equation 4.
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