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.
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.
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