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Homework Help: Scalar Equations: How to solve them

  1. Feb 26, 2010 #1
    My name is Abhishek Agrawal. I am a grad student in Computer Science. I know very little about scalar equations. But my research has led me to them and I am at my wits end on how to go about solving them. I have looked around every where but I am unable to start.

    1. The problem statement, all variables and given/known data


    This is a Wikipedia link on how to use the divide and conquer algorithm to solve for eigen values of a matrix. I understood most of the stuff but towards the end of the link, the author ends up with this equation

    1 + wT(D − λI)−1 w = 0

    which is a scalar equation. He transforms this equation into an another form. (The formatting came bad on this post, better to view it in Wiki)

    1 + \sum_{j=1}^{m} \frac{w_{j}^{2}}{d_{j} - \lambda} = 0.

    What I cannot understand is how is the summation applied. How does the author get the wj and dj terms from the matrix . Is he using an entire row or is there some thing I am missing ?

    I would appreciate if some one could give me some links so as to what I am missing or could point me to some tutorials
    Any help would be appreciated
  2. jcsd
  3. Feb 27, 2010 #2


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    Homework Helper

    Hint: if you wrap the TeX code in tex tags (type [ tex ] without the spaces, then this code, then [ / tex ] without the spaces - ie [/tex]), then you will get a nice formatting, like so:
    [tex]1 + \sum_{j=1}^{m} \frac{w_{j}^{2}}{d_{j} - \lambda} = 0. [/tex]

    Answer to your question: note that D is a diagonal matrix. Therefore, D - λI is a diagonal matrix as well, looking like
    [tex]D - \lambda I = \begin{pmatrix} d_1 - \lambda & & 0 \\ & \ddots & \\ 0 & & d_m - \lambda \end{pmatrix}[/tex]

    The inverse of this matrix is then simply
    [tex](D - \lambda I)^{-1} = \begin{pmatrix} 1/( d_1 - \lambda ) & \cdots & 0 \\ \vdots & \ddots & \\ 0 & & 1 / (d_m - \lambda) \end{pmatrix}[/tex]

    If you now perform the matrix multiplication explicitly for a vector w = (w1, ..., wm) you will find the second expression.
  4. Feb 27, 2010 #3
    Thank you for your quick reply. I need one more clarification.
    When the author writes
    1 + \sum_{j=1}^{m} \frac{w_{j}^{2}}{d_{j} - \lambda} = 0.

    in this equation [tex]d_{j} [/tex], does this mean [tex]d_{j,j} [/tex] ?
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