Canonical form and change of coordinates for a matrix

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jejaques
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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 [tex]\lambda[/tex] on the diagonal, take the determinant, find the roots by solving for the [tex]\lambda[/tex] values, and then plug them in one at a time to find the different [tex]\zeta[/tex], 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 [tex]\lambda[/tex] 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!
 
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Welcome to PF!

Hi jejaques! Welcome to PF! :smile:

(have a lambda: λ :wink:)
jejaques said:
Unfortunately, my professor has only shown us the various [tex]\lambda[/tex] 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 :smile:
 


tiny-tim said:
Hi jejaques! Welcome to PF! :smile:

(have a lambda: λ :wink:)


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


tiny-tim said:
Hi jejaques! Welcome to PF! :smile:

(have a lambda: λ :wink:)


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


Hello, and thanks for the welcome...

Yeah, my reasoning was in my "attempt at a solution" section. I subtracted [tex]\lambda[/tex] 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 [tex]\lambda[/tex]6 - 2[tex]\lambda[/tex]3 + 1

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

It has identical real roots at... [tex]\lambda<sub>1</sub>[/tex] = [tex]\lambda<sup>2</sup>[/tex] = 1, and identical complex roots at [tex]\lambda<sub>3</sub>[/tex] = [tex]\lambda<sub>4</sub>[/tex] = 1/2 + [tex]\sqrt{3}[/tex]i[tex]/[/tex]2 and [tex]\lambda<sub>5</sub>[/tex] = [tex]\lambda<sub>6</sub>[/tex] = 1/2 - [tex]\sqrt{3}[/tex]i[tex]/[/tex]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[tex]\bar{}[/tex]:
[tex]\alpha[/tex] [tex]\beta[/tex] 0
-[tex]\beta[/tex] [tex]\alpha[/tex] 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 [tex]\lambda<sub>1</sub>[/tex] = [tex]\lambda<sub>2</sub>[/tex] canonical form in that big 6 x 6?

Thanks!