Question

is there any rule to find the characteristic equation of 4x4 matrices?

Answer 1

nah just eigenvalues that I know of but for reducing it helps to make it diagonal or upper triangular

Answer 2

the first step to get diagonalized a matrix is make up characteristic equation. I ask step by step

Answer 3

I know how to do with 2x2 and 3x3 not 4x4

Answer 4

unfortunately it is alot of annoying calculations

Answer 5

same idea, you know how to get the det of a 4x4?

Answer 6

yes, but it is much more simple when getting det of 4x4 by using rref. than cofactor method

Answer 7

eh, yea I mean the beauty of it is that if you have an upper/lower triangular or diagonal, it's pretty easy but other than that it is a lot of useless computations
I assume you know the formula for the characteristic eq to be \[det(\lambda I-A)=0\]

Answer 8

i know how to diagonalize a 3x3 matrix and the formula to get the characteristic equation from the original matrix without using much calculation. I wonder whether there is some method I can apply to 4x4 or not.

Answer 9

anyway, thanks a lot. let me try to ask my pro

Answer 10

do you know that formula ? i mean formula to get 3x3 characteristic equation

Answer 11

I just apply the one I told you, but I can't remember if putting it in ref is allowed or not

Answer 12

that would be my choice and it doesn't change my answer in the long run... I don't think

Answer 13

it's L^3 - (trace)L^2 + (A11+A22+A33)L -det (A) =0

Answer 14

L is lamda

Answer 15

We never covered trace, my teacher deemed it unimportant so I just take the long way i guess

Answer 16

but that is the form of a 3x3

Answer 17

no, mine is not from myself, from a very experience professor. and that works perfectly

Answer 18

sure. so that's why i am looking for another perfect for 4x4

Answer 19

hmm so very interesting

Answer 20

well it will always be of the form \[\lambda^{n}+c_{1}\lambda^{n-1}+...+c_n=0\]

Answer 21

yes. I know that formula

Answer 22

the last term is det. the coefficient of second one is trace

Answer 23

because the eigenvalues must be unique we can assume the degree of the polynomial is equal to that of the matrix

Answer 24

ok. I need more time to figure out what I have to know.

Answer 25

but your answer from there, I can neither confirm nor deny

Answer 26

midterm?

Answer 27

got it. I will ask my pro. if I have any thing interesting, i will let you know

Answer 28

no. not that. just wonder

Answer 29

I am a crazy learner.

Answer 30

lol curiosity is a good thing

Answer 31

I'm on here to brush up, I actually have linear algebra tomorrow, and this was a nice refresher for eigenvalues

Answer 32

me too, I have linear and discrete tomorrow. i have to go to bed now. too late. see you tomorrow

Answer 33

have fun

Answer 34

good night. friend

Answer 35

you as well

Answer 36

hey , anyone have any idea but det. please

Answer 37

tri-diagonal form will reduce the number of computations involed.
check LINPACK on internet. you'll find out

Answer 38

thanks electrokid