Re: Finding eigenvalues

Arturo Magidin wrote/skrev/kaita/popisal/schreibt :
Konrad Viltersten <tmp1@xxxxxxxxxxxxxx> wrote:
We have the follownig matrix

Obviously, you do have a functioning shift key, since you
were able to write an upper case W.

A = [0 1 0 ; 0 0 1 ; 1 0 0]
and when i try to determine it's eigenvalues i get the

so then why do you not use it for the "I", the way it's
supposed to?

Maybe for the same reason as you wrote "a polynomial... have
n roots" instead of "a polynomial... _HAS_ n roots". Surely,
you do have the notion of plural form in your language... :)

Now, to make sure we're friends here so nobody will start
a flame war - i'm always polite, always say "thanks", never
pick a fight, never bull etc. I think the group will manage
me not using capitals for pronouns, hehe.

I hope you don't take offence i do appologize if you do.

equation of 1 - lambda^3 = 0 , which i then solve and
recieve lambda = 1 as the only root.

Over the real numbers, yes. The only real root of
lambda^3-1 = 0 is lambda=1.

Now, for some
reason MatLab eagorly tells me that there are three
different eigenvalues, two of which are complex.

Well, yes. A polynomial of degree n always have n roots over the
complex number (counting multiplicity).

How does it know it?

In this case, trivially. Factor 1-lambda to get
(1-lambda^3) = (1-lambda)(1 + lambda + lambda^2)
and then solve the quadratic lambda^2 + lambda + 1 using any method,
e.g., the quadratic formula, to get that the other two (complex) roots
are
(-1 + sqrt(-3))/2 and (-1-sqrt(-3))/2
the two other cubic roots of 1 in the complex numbers.


Right, got it. I was too focused on the reals. Thanks!

--
Vänligen
Konrad
---------------------------------------------------

Sleep - thing used by ineffective people
as a substitute for coffee

Ambition - a poor excuse for not having
enough sence to be lazy
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.

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