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  1. MHB Master
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    #1
    I am reading the book: Multivariable Mathematics by Theodore Shifrin ... and am focused on Chapter 8, Section 2, Differential Forms ...

    I need some help in order to fully understand some statements of Shifrin at the start of Chapter 8, Section 2 on the dual space ...

    The relevant text from Shifrin reads as follows:









    In the above text from Shifrin we read the following:

    " ... ... Then $ \displaystyle \phi = a_1 dx_1 + \ ... \ ... \ + a_n dx_n$ ... ... "



    Can someone please demonstrate and explain how/why $ \displaystyle \phi = a_1 dx_1 + \ ... \ ... \ + a_n dx_n$ ... ...



    Help will be much appreciated ... ...

    Peter

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    #2
    Hi Peter,

    To establish the result, let $v=\sum_{k=1}^{n}v^{k}e_{k}$ be a vector in $\mathbb{R}^{n}$. Then, using linearity of $\phi$ judiciously, $$\phi(v) = \sum_{k=1}^{n}v^{k}\phi\left(e_{k}\right)=\sum_{k=1}^{n}a_{k}v^{k}.$$ On the other hand, using linearity of the $dx^{k}$,
    \begin{align*}
    \left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(v\right) &= \left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
    &= a_{1}dx^{1}\left(\sum_{k=1}^{n}v^{k}e_{k}\right) +\cdots + a_{n}dx^{n}\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
    &=a_{1}v^{1}+\cdots +a_{n}v^{n}.
    \end{align*}
    Since the two are equal and $v$ was chosen arbitrarily, we have $\phi=a_{1}dx^{1}+ \cdots + a_{n}dx^{n}.$

  4. MHB Master
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    #3 Thread Author
    Quote Originally Posted by GJA View Post
    Hi Peter,

    To establish the result, let $v=\sum_{k=1}^{n}v^{k}e_{k}$ be a vector in $\mathbb{R}^{n}$. Then, using linearity of $\phi$ judiciously, $$\phi(v) = \sum_{k=1}^{n}v^{k}\phi\left(e_{k}\right)=\sum_{k=1}^{n}a_{k}v^{k}.$$ On the other hand, using linearity of the $dx^{k}$,
    \begin{align*}
    \left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(v\right) &= \left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
    &= a_{1}dx^{1}\left(\sum_{k=1}^{n}v^{k}e_{k}\right) +\cdots + a_{n}dx^{n}\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
    &=a_{1}v^{1}+\cdots +a_{n}v^{n}.
    \end{align*}
    Since the two are equal and $v$ was chosen arbitrarily, we have $\phi=a_{1}dx^{1}+ \cdots + a_{n}dx^{n}.$

    Thanks GJA ... most helpful ...

    Appreciate your help ...

    Peter

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