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Orthogonal Eigenvector, Proof is bothering me

  1. Jun 29, 2008 #1
    Suppose

    [tex]A\overrightarrow{x}=\lambda_1\overrightarrow{x}[/tex]
    [tex]A\overrightarrow{y}=\lambda_2\overrightarrow{y}[/tex]
    [tex]A=A^T[/tex]

    Take dot products of the first equation with [tex]\overrightarrow{y}[/tex] and second with [tex]\overrightarrow{x}[/tex]

    ME 1) [tex](A\overrightarrow{x})\cdot \overrightarrow{y}=(\lambda_1\overrightarrow{x})\cdot\overrightarrow{y}[/tex]

    BOOK ... skipped steps but only shows this 1) [tex](\lambda_1\overrightarrow{x})^T\overrightarrow{y}=(A\overrightarrow{x})^T\overrightarrow{y}=\overrightarrow{x}^TA^T\overrightarrow{y}=\overrightarrow{x}^TA\overrightarrow{y}=\overrightarrow{x}^T\lambda_2\overrightarrow{y}[/tex]

    Now it looks like I have to transpose my first step, but if I do so, do I assume that [tex]y=y^T[/tex]?
     
    Last edited: Jun 29, 2008
  2. jcsd
  3. Jun 30, 2008 #2

    CompuChip

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    Note that the dot product of two vectors is just a matrix multiplication. E.g. if we take [itex]\vec x, \vec y[/itex] to be (n x 1) column vectors then the dot product is just
    [tex]\vec x \cdot \vec y = \vec x^T \vec y[/tex]
    where the right hand side is just matrix multiplication of an (1 x n) with an (n x 1) matrix. Then the solution easily follows. Try to write it out and indicate what is being done in each step.
     
  4. Jun 30, 2008 #3

    HallsofIvy

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    You don't, they are not the same.

    What you are doing is thinking of the dot product [itex]\vec{u}\cdot\vec{v}[/itex] as the matrix product [itex]\vec{u}^T\vec{v}[/itex] where [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] are "column" matrices and [itex]\vec{u}^T[/itex] is the "row" matrix corresponding to [itex]\vec{v}[/itex]. It Then follows that [itex](A\vec{x}\cdot\vec{y}= (A\vec{x})^T\vec{y}= \vec{x}^TA^T\vec{y}[/itex] which, because AT= A is the same as [itex]\vec{x}^TA\vec{y}= \vec{x}\cdot A\vec{y}[/itex]. It is not that [itex]\vec{y}^T= \vec{y}[/itex], you never have "[itex]\vec{y}^T[/itex].
     
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