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So I'm doing Inverse Iteration in Excel to find the dominant eigenvalue and eigevector of a matrix.

This particular method involves estimating an eigenvalue, multiplying the identity matrix by it, then subtracting from the original matrix.

From this point, the iteration begins by multiplying the inverse matrix with an intial vector

$v_0 = $

$ \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} $

to find $u_1$

Now the next step is normally ok, but in this case it confused me

Normally you would normalise $u_1$ by dividing by the maximum value in the vector, for example -

$u_1 = $

$ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} $

Then you would normalise by dividing the entire vector by 3

However in this case

$u_1 =$

$ \begin{matrix} 0.181818 \\ -0.66667 \\ 0.109091 \\ \end{matrix} $

The method says that the maximum value of this vector is the -0.66667, why is this the case? Is it always the case that you look at the number regardless of the sign to determine the maximum value?

Let me know if you need any clarifications.

NMH
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  • Let me know if it's not clear exactly what I'm asking and I'll try to explain. – NMH Jun 21 '14 at 21:12
  • the dominant eigenvalue is the one that is greatest in magnitude, so you take the absolute value. – bobbym Jun 21 '14 at 23:30
  • Ah, I see, so the sign becomes irrelevant. But then why is it not divided by the absolute value to normalise? – NMH Jun 22 '14 at 18:28
  • It is frowned upon to continue to post comments back and forth so this is my last. As I understand it, this normalization is used to keep the vector entries small, otherwise they would blow up causing problems in calculation. I have a book here where he picks the smallest value and divides by that?! I would choose the largest one in absolute value, this makes sure that you never divide by 0. I think you could divide by the -.66667.

    I have also seen using the distance norm of the vector which would always be positive to normalize. Sorry, if that does not clear the problem up a little.

     – bobbym Jun 22 '14 at 22:21
  • The common approach to normalize the vectors in the power method is to divide the vector by its $\infty$-norm, which is that maximal magnitude of its entries. This is to avoid underflow/overflow in the finite precision arithmetic. – Algebraic Pavel Jun 23 '14 at 15:55

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