# Canonical form and change of coordinates for a matrix

1. Mar 25, 2009

### jejaques

Hello! I'm trying to do some linear algebra. I have an insane Russian teach whose English is, uh, lacking.. so I'd appreciate any help with these I can get here!

1. The problem statement, all variables and given/known data
Find the canonical forms for the following linear operators and the matrices for the corresponsing change of coordinates.

Here is the 6x6 matrix:
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
-1 0 0 -2 0 0

2. Relevant equations

3. The attempt at a solution
I know I have to do subtract $$\lambda$$ on the diagonal, take the determinant, find the roots by solving for the $$\lambda$$ values, and then plug them in one at a time to find the different $$\zeta$$, turn that into a change of coordinates, and then depending on case, put it into canonical form...

Unfortunately, my professor has only shown us the various $$\lambda$$ cases for 2 x 2 matrices and because we can "look everything up on google," we have no book!!!

A couple questions: Can I simplify this or maybe turn it into the Jordan block? Can anyone point me to a similar problem, even? I've been searching for two hours, have searched through three free linear algebra e-books and am still lost =(

Thanks so much!

2. Mar 26, 2009

### tiny-tim

Welcome to PF!

Hi jejaques! Welcome to PF!

(have a lambda: λ )
Just put -λ down the diagonal, and calculate the determinant

3. Mar 26, 2009

### jejaques

Re: Welcome to PF!

Hello, and thanks for the welcome...

Yeah, my reasoning was in my "attempt at a solution" section. I subtracted $$\lambda$$ from the diagonal and did the determinant; I just thought it was too much tedious stuff to post here, as I'm having problems further on.

The determinant is $$\lambda$$6 - 2$$\lambda$$3 + 1

To factor roots, I set the determinant equal to zero and factored, as follows:
0 = ($$\lambda$$3 - 1)2
= ($$\lambda$$-1)($$\lambda$$5 + $$\lambda$$4 + $$\lambda$$3 - $$\lambda$$2 - $$\lambda$$ - 1)
= ($$\lambda$$ - 1)($$\lambda$$ - 1)($$\lambda$$4 + 2$$\lambda$$3 + 3$$\lambda$$2 + 2$$\lambda$$ + 1)
= ($$\lambda$$ - 1)2($$\lambda$$2 + $$\lambda$$ + 1)2

It has identical real roots at... $$\lambda1$$ = $$\lambda2$$ = 1, and identical complex roots at $$\lambda3$$ = $$\lambda4$$ = 1/2 + $$\sqrt{3}$$i$$/$$2 and $$\lambda5$$ = $$\lambda6$$ = 1/2 - $$\sqrt{3}$$i$$/$$2

But the issue is, with a 6 x 6 matrix, which case should I evaluate and how should I go about finding the eigenvectors?

I know complex roots evaluate to the canonical form A$$\bar{}$$:
$$\alpha$$ $$\beta$$ 0
-$$\beta$$ $$\alpha$$ 0
0 0 1

But do I need to evaluate each of the positive and negative complex roots separately, and where do I throw in the $$\lambda1$$ = $$\lambda2$$ canonical form in that big 6 x 6?

Thanks!

4. Apr 29, 2009

### msakaria

Hey, buddy are you in sergey nikitin class at ASU?