This is one of the templates used in this class because it has some shortcuts that make it easy to compute by hand.

The λ values can always be found with a few tricks. First, we need to pseudo-condense it to a form we can work with easier.

by letting A=α+iβ and B=a+ib in complex form.

Now, we can follow our general rules:

λ₁=A+B(n-1)= (α + a)+(β + b)i=α₁ +iβ₁

λ₂ = A-B=(α - a)+(β - b)i=α₂ +iβ₂

This gives us only two λ values but we can expand it back to the original matrix which needs four λ values by adding the complex conjugates. However, to use the C matrix described below you must use the first two λ values calculated regardless of the sign of the imaginary component.

Example 1

Pseudo-Condensed

λ₁=A+B(n-1)= (α + a)+(β + b)i=α₁ +iβ₁= 5i

λ₂ = A-B=(α - a)+(β - b)i=α₂ +iβ₂= -2-i

To use the C matrix below you must use the first two λ values calculated regardless of the sign of the imaginary component.

Example 2

Pseudo-Condensed

λ₁=A+B(n-1)= (α + a)+(β + b)i=α₁ +iβ₁= 7-3i

λ₂ = A-B=(α - a)+(β - b)i=α₂ +iβ₂= 5+7i

To use the C matrix below you must use the first two λ values calculated regardless of the sign of the imaginary component.

Note: with software available you can check to see if you came up with a correct solution by evaluating C^-1*A*C and if it gives you your A_new you are correct.

For: Sergey Nikitin

MAT 343 FALL 2010

By:Joshua Carroll and Carolina Tostado