What is the Correct Way to Find Eigenvalues and Eigenvectors of a Matrix?

Seon
Messages
1
Reaction score
0
Find the eigenvalues and corresponding eigenvector of the matrix.
A=
[-4 4 8 ]
[0 0 -10]
[0 0 2 ]

[1 -1 0]
~ [0 0 1 ]
[0 0 0 ]

I calculated by A = -\lambdaI

So,

[1-lamda -1 0 ]
[0 -lamda 1]
[0 0 -lamda]

so, lamda = 0,0, and 1

So I got

1st eigen value: 0 eigen vector (1,1,0)
2nd eigen value: 0 eigen vector (1,1,0)
3rd eigen value: 1 eigen vector (1,0,0)

1st and 2nd values were right, but third one was wrong.
I tried several times, and I always get 1(1,0,0)

What do i need to do ?
thanks
 
Physics news on Phys.org
if you reduce the matrix, you change the eigenvalues, except for 0. don't reduce the matrix, find the characteristic polynomial of the original A.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top