
#1
Aug2311, 09:44 AM

P: 24

Hello i have this matrix [itex]\in Z [/itex] mod [itex] 7[/itex],
M = \begin{pmatrix} 0&6\\ 5&0 \end{pmatrix} always modulo [itex]7[/itex] in [itex]Z[/itex]. I found characteristic polynomial [itex]x^2+5[/itex]. Eigenvalues are [itex]\lambda = 3, \lambda' = 4[/itex] Eigenvectors related to [itex]\lambda = 3 [/itex] are the nonzero solution of the system: [itex]4x +6y = 0,[/itex] [itex]5x+4y = 0 [/itex] I get: [itex]4x = y,[/itex] [itex]6y[/itex] I don't know if it is correct, but how can i find the eigenvectors? 



#2
Aug2311, 10:00 AM

Mentor
P: 16,512

Yes, y=4x is correct. Now you just need to find a nonzero vector (0,0) that satisfies this. So, we can take y=1 for example. In that case (x,y)=(4,1) satisfies the equation. So (4,1) is an eigenvector.
We could also take y=2, then y=(1,2) was an eigenvector. etc. 



#3
Aug2411, 03:15 AM

P: 24

now i have a problem to solve this:
in [itex]Z[/itex] mod [itex]11[/itex] M = \begin{pmatrix} 0&2\\ 7&0 \end{pmatrix} The characteristic polinomyal is : [itex]x^2+8[/itex], eigenvalues [itex]\lambda=5, \lambda'=6[/itex], eigenvectors related to [itex]\lambda=5[/itex] are the nonzero solution of the system: [itex]6x+2y=0[/itex] [itex]7x+6y=0[/itex] I don't know how to solve this system because i took a look at the solution of the exercise and it is: [itex]V={(y, 8y)}[/itex] I don't know how to get this solution. Oh i get now i'm wrong to write the first equation. 



#4
Aug2411, 06:40 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

eigenvalues and eigenvectors of a matrix 


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