Canonical form and change of coordinates for a matrix

[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!

Homework Statement

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

Homework Equations

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!

 

[tiny-tim]

Welcome to PF!

Hi jejaques! Welcome to PF!

(have a lambda: λ )

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!

Just put -λ down the diagonal, and calculate the determinant

 

[jejaques]

Hi jejaques! Welcome to PF!

(have a lambda: λ )


Just put -λ down the diagonal, and calculate the determinant

Hi jejaques! Welcome to PF!

(have a lambda: λ )


Just put -λ down the diagonal, and calculate the determinant

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 \lambda⁶ - 2\lambda³ + 1

To factor roots, I set the determinant equal to zero and factored, as follows:
0 = (\lambda³ - 1)²
= (\lambda-1)(\lambda⁵ + \lambda⁴ + \lambda³ - \lambda² - \lambda - 1)
= (\lambda - 1)(\lambda - 1)(\lambda⁴ + 2\lambda³ + 3\lambda² + 2\lambda + 1)
= (\lambda - 1)²(\lambda² + \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!

 

[msakaria]

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