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i mean det (A -(lamda)I)= 0 say for example A = (1 2;-1 4)

i know the eigen values are lamda = 3 and lamda =2 but how to find eigenvectors

To be honest, I'm not too great at eigenvectors either.

You should be able to use any numbers you want, unless I'm grossly misunderstanding eigenvectors.

They wanted a normalized vector apparently (length of 1). Anyways, I've got to get going. Good luck.

ooo alright thanks a lot though

but @DanielxAK to get a normalized eigen vector we find the main eigenvector and normalize it?

A bit. What's your question?

is this like a spectral decomposition?

hmm not quite

i didnt know (C*DC)* = (C*)*D*C* like i thought it was (AB)* = B*A* ?

hmm true do you mind showing me an example of spectral decomposition if you dont mind ?

i have A = ( 3 6 -1; 6 9 4; -1 4 3)

ok sounds good and C= the normalized matrix for A?

ahh okayy i have to keep on practicing thenn i apparently have to do this by hand tommorow lol

I have A = (3 1 1; 1 0 2; 1 2 0)

and i found the eigenvalues to be lamda1=1, lamda2=4, lamda3 = -2