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Eigenvalues and eigenvectors of a matrix

by blob84
Tags: eigenvalues, eigenvectors, matrix
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blob84
#1
Aug23-11, 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 non-zero 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?
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micromass
#2
Aug23-11, 10:00 AM
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Yes, y=4x is correct. Now you just need to find a non-zero 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.
blob84
#3
Aug24-11, 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 non-zero 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.

HallsofIvy
#4
Aug24-11, 06:40 AM
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Sci Advisor
Thanks
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P: 39,339
Eigenvalues and eigenvectors of a matrix

Quote Quote by micromass View Post
Yes, y=4x is correct. Now you just need to find a non-zero 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.
You have this backwards. If y= 4x, then the eigenvector is (1, 4), not (4, 1).


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