matrix eigenvector Watch

ellishnoo
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#1
Report Thread starter 7 years ago
#1
Having problems understanding eigenvectors,
working on 3x3 matrix,
(4 0 1)
(-1 -6 -2)
(5 0 0) (sorry) dont know how to do a matrix in latex.

I can find the characteristic polynomial.
but when it comes to finding the eigenvectors I keep getting confused..
 

-10x + 0y -    z = 0  .eq 1

   - x + 0y + 2z = 0  .eq 2

-  5x + 0y -  6z = 0  .eq 3

the above problem is when lambda equals -6,
I know the answer is meant to be 0 1 0.. but cant work out how to do it..

ive seen 2 methods which result in similar answer but dont understand process..

either using replacing say colum z with t so youd have

 

-10x  = t  .eq 1

   - x  = -2t , then ....  x = 2t  .eq 2

-  5x = 6t    .eq 3

how does that get to x,y,z,=0,1,0
this would imply that the the y colum was seen as t, making

-10x + 0t - z = 0 .eq 1
- x + 0t + 2z = 0 .eq 2
- 5x + 0t - 6z = 0 .eq 3

t would vanish... how does column x and z translate as zero???.
im going a bit round in circles, any guidance, as all Ive seen so far do not explain the step quite often they look at the resultant matrix, and then say obviosuly the vectors are such and such, it seems to imply there is something simple I am missing that my teacher has skipped to tell us.

as always any help greatly appreciatted

Cheers elli
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User592005
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I'm sorry for being so direct, but this thread is useless if you don't state the original question.
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ellishnoo
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(Original post by tulcod)
I'm sorry for being so direct, but this thread is useless if you don't state the original question.
thanks for the heads up, ive put the origianl matrix in my original question.
thanks
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User592005
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(Original post by ellishnoo)
thanks for the heads up, ive put the origianl matrix in my original question.
thanks
so what's the original question?
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ellishnoo
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(Original post by tulcod)
so what's the original question?
find the eigenvector of said matrix
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User592005
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for lambda=-6, the matrix becomes:
10 0 1
-1 0 -2
5 0 6
it's slightly different from what you gave, but it doesn't give an eigenvector of 0 1 0. perhaps that's for a different eigenvalue?
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ellishnoo
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(Original post by tulcod)
for lambda=-6, the matrix becomes:
10 0 1
-1 0 -2
5 0 6

it's slightly different from what you gave, but it doesn't give an eigenvector of 0 1 0. perhaps that's for a different eigenvalue?
I used several online caluclators and they all gave the same answer, 0 1 0..??
i used row echelon form
and broke it down to

[1 0 0][x]=0
[0 0 1][y]=0
[0 0 0][z]=0

but this breaks down to
x=0, y=0, z= doestnt exist or maybe you could call it one, I dont know why??

I know there are two methods to find the eigenvalues, I have been informed, that if A and I are n x n matrices, then \det(A-\lambda I) = (-1)^n \det (\lambda I - A). In other words one is only the minus of the other when n is odd.

so that is why we have different signs prob, I hope

So any ideas how to find the eigenvectors of the above reduced form matrix, if its the correct or sort of correct form
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