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I Bilinear forms

  1. Jul 3, 2017 #1
    I'm having trouble understanding a step in a proof about bilinear forms
    Let ## \mathbb{F}:\,\mathbb{R}^{n}\times\mathbb{R}^{n}\to \mathbb{R}## be a bilinear functional.
    ##x,y\in\mathbb{R}^{n}##
    ##x=\sum\limits^{n}_{i=0}\,x_{i}e_{i}##
    ##y=\sum\limits^{n}_{j=0}\;y_{j}e_{j}##
    ##F(x,y)=F(\sum\limits^{n}_{i=1}\,x_{i}e_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j})=\sum\limits^{n}_{1=i,j}F(x_{i}e_{i},y_{j}e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}F(e_{i},e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}a_{ji}##
    ##=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}##

    Could someone explain to me how we got that
    ##F(e_{i},e_{j})=a_{ji}##
    I couldn't find the explanation anywhere in my notebook
    thank you
    Edit: ( added some stuff I missed )

    ##A=[a_{ij}]_{i,j=1,...,n}##

    ##<Ax,y>=<A(\sum\limits^{n}_{i=1}x_{i}e_{i}),\sum\limits^{n}_{j=1}y_{j}e_{j}>=\sum\limits^{n}_{i=1}x_{i}Ae_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j}>=\sum\limits^{n}_{i,j=1}x_{i}y_{j}<Ae_{i},e_{j}>##

    ##=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}##

    I understand why ##\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}=<Ax,y>##,
    however I dont understand why ##<Ax,y>=F(x,y)##
    Edit(Edit).
    In other words I don't understand why there exists a Matrix A so that ##F(x,y)=<Ax,y>##
     
    Last edited: Jul 3, 2017
  2. jcsd
  3. Jul 3, 2017 #2

    Orodruin

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    ##F(e_i, e_j)## is a number. Define that number to be ##a_{ij}##.
     
  4. Jul 3, 2017 #3

    StoneTemplePython

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    I assume that ##\mathbf e_k## refers to standard basis vectors, i.e.

    ##\mathbf I =
    \bigg[\begin{array}{c|c|c|c}
    \mathbf e_1 & \mathbf e_2 &\cdots & \mathbf e_{n}
    \end{array}\bigg]##

    I would start real simple. Suppose we had the constraint that ##\mathbf x := \mathbf e_k## and ##\mathbf y := \mathbf e_r##.

    What happens when you do ##\mathbf e_r^T \mathbf A \mathbf e_k = \mathbf y^T \mathbf A \mathbf x##? This equation "grabs" one cell from ##\mathbf A##, which you should verify by inspection.

    From here recall that the standard basis vectors in fact form a basis, so let's loosen up our restriction and let ##\mathbf x## be anything (while keeping its dimension and field the same, of course.)

    ##\mathbf x = \gamma_1 \mathbf e_1 + \gamma_2 \mathbf e_2 + ... + \gamma_n \mathbf e_n##

    ##\mathbf y^T \mathbf A \mathbf x = \mathbf y^T \mathbf A\big(\gamma_1 \mathbf e_1 + \gamma_2 \mathbf e_2 + ... + \gamma_n \mathbf e_n\big)##

    now loosen up the constraint on ##\mathbf y## so it too can be anything

    ##\mathbf y = \eta_1 \mathbf e_1 + \eta_2 \mathbf e_2 + ... + \eta_n \mathbf e_n##

    ##\mathbf y^T \mathbf A \mathbf x = \big(\eta_1 \mathbf e_1 + \eta_2 \mathbf e_2 + ... + \eta_n \mathbf e_n \big)^T \mathbf A\big(\gamma_1 \mathbf e_1 + \gamma_2 \mathbf e_2 + ... + \gamma_n \mathbf e_n\big) = \sum \sum \eta_j \gamma_i a_{j,i} = \sum \sum \eta_j \gamma_i F(e_{i},e_{j}) = \sum \sum y_j x_i F(e_{i},e_{j}) ##

    If you work through that, any bi-linear form your want (over reals with finite dimension) is just "grabbing" and scaling stuff from ##\mathbf A## and then summing them up so you get a scalar result.
     
  5. Jul 3, 2017 #4

    WWGD

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    If A is linear, then <Ax,y> is bilinear and so it has a representation as a matrix, as a quadratic form. Is that your question?
     
  6. Jul 3, 2017 #5

    StoneTemplePython

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    follow-up thought:

    my posting had ##\mathbf y^T \mathbf {Ax} = \mathbf y^T\big( \mathbf {Ax}\big) = <\mathbf y, \mathbf{Ax}>##

    though OP actually asked for

    ##< \mathbf{Ax}, \mathbf y>##

    inner products over Reals are symmetric, so

    ##< \mathbf{Ax}, \mathbf y> = <\mathbf y, \mathbf{Ax}>##

    (over complex numbers inner products are symmetric, plus complex conjugation at the end)

    note that Chapter 7 of Linear Algebra Done Wrong has a very good discussion of bilinear and quadratic forms. Freely available by the author here:

    https://www.math.brown.edu/~treil/papers/LADW/book.pdf
     
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