# Scalar Equations: How to solve them

1. Feb 26, 2010

### abhimanipal

Hi,
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

http://en.wikipedia.org/wiki/Divide-and-conquer_eigenvalue_algorithm

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. Feb 27, 2010

### CompuChip

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

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

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

If you now perform the matrix multiplication explicitly for a vector w = (w1, ..., wm) you will find the second expression.

3. Feb 27, 2010

### abhimanipal

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 $$d_{j}$$, does this mean $$d_{j,j}$$ ?

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