# Characteristic Polynomial from a matrix with an unknown...

• Nov 10th 2007, 06:06 AM
sh01by
Characteristic Polynomial from a matrix with an unknown...
Hi, for the second time today, still linear algebra...

Okay, if I have the matrix,

A = [2a a 0]
--- [ a a 0]
--- [ 0 0 a]

and I want to find the eigenvalues of the matrix, I know I need to set up the characteristic polynomial, which I find to be...

(2a-t)(a-t)^2 - a^2(a-t)

I have made various adjustments, but I still haven't found the roots of the polynomial.

What I have come up with is either...

-t^3 + 4at^2 - 4a^2t - a^3

or

(a-t)(t^2 - 3at + a^2)

Okay, so... No matter which equation I look at, I still can't find the roots.
I can see from the last one -- that one root should be t = a? But division comes up with an illegal answer.

So, if anyone has any input, it is very welcome.

Thanks.
Simon DK.
• Nov 10th 2007, 06:27 AM
kalagota
Quote:

Originally Posted by sh01by
Hi, for the second time today, still linear algebra...

Okay, if I have the matrix,

A = [2a a 0]
--- [ a a 0]
--- [ 0 0 a]

and I want to find the eigenvalues of the matrix, I know I need to set up the characteristic polynomial, which I find to be...

(2a-t)(a-t)^2 - a^2(a-t)

I have made various adjustments, but I still haven't found the roots of the polynomial.

What I have come up with is either...

-t^3 + 4at^2 - 4a^2t - a^3

or

(a-t)(t^2 - 3at + a^2)

Okay, so... No matter which equation I look at, I still can't find the roots.
I can see from the last one -- that one root should be t = a? But division comes up with an illegal answer.

So, if anyone has any input, it is very welcome.

Thanks.
Simon DK.

what do you mean by ".. But division comes up with an illegal answer."

yes, one eigenvalue must be t=a. for the other two, solve using quadratic formula.. i just wonder if a is unknown or just an arbitrary variable from you field..
• Nov 10th 2007, 06:37 AM
sh01by
I tried to divide (t-a) with (t^2 - 3at + a^2), but in the end I had a^2 left, so I came up with no solution. I have tried once again to use the quadratic formula -- and this time the answer seems more plausible... I have tried to set a to 5 and plot it on my ti-83 and now the roots are okay. Strange how I found it to be wrong the first time.

Thanks. You're quite active here. (Handshake)
:)
• Nov 10th 2007, 06:41 AM
kalagota
hehe.. if you divide your quadratic equation by a and got a remainder, then a must be a unique eigenvalue of your matrix..

hmm, for some reasons, i am active because it saturday here and i have a lot of time checking the site.. but during weekdays, i am in school almost the whole day that is why when i come back home, it is already late and i have to sleep for tomorrow.. Ü