The matrix below gives me 3 eigen values. 2 which are equal:
[2 0 0]
[1 3 0 ]
[1 1 2 ]
When i type this matrix into Eigen value calculator online it says the corresponding Eigen vectors are: For the eigen value 2, i get (1,-1,0), (0,0,1)
For the eigen value 3, i get: (0,1,1)
how does the calculator know the alternative eigen vector is (0,0,1)? because if i sub 2 into det(I-A*Lambda) i will definately get same eigen vector again, dont understand how calculator got
Does that mean that, i would still be correct if i repeated the eigen vector? or do i have to give the alternative representation like the calculator (0,0,1)
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Basically what i am trying to do is diagnolisation, and i need to find the eigen vectors and join them as 1 matrix.
[2 0 0]
[1 3 0 ]
[1 1 2 ]
When i type this matrix into Eigen value calculator online it says the corresponding Eigen vectors are: For the eigen value 2, i get (1,-1,0), (0,0,1)
For the eigen value 3, i get: (0,1,1)
how does the calculator know the alternative eigen vector is (0,0,1)? because if i sub 2 into det(I-A*Lambda) i will definately get same eigen vector again, dont understand how calculator got
Does that mean that, i would still be correct if i repeated the eigen vector? or do i have to give the alternative representation like the calculator (0,0,1)
------------------
Basically what i am trying to do is diagnolisation, and i need to find the eigen vectors and join them as 1 matrix.
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These vectors are obtained from the nullspace of
0 0 0
1 1 0
1 1 0
the eigenvectors corresponding to 2 are those which are in the nullspace of the above matrix.
If we set x3=0, the non-trivial result eigenvector has x1=-x2... for simplicity, we say x1=1
==> x2=-1
hence the first eigenvector is (1,-1,0)
Next, set x3=1, we can now have x1=x2=0 without obtaining a trivial eigenvector, so the second eigenvector is (0,0,1) (although, you could also use non-zero x1 and x2, but this is the simplest case which preserves linear independence).
I hope this has helped :)
0 0 0
1 1 0
1 1 0
the eigenvectors corresponding to 2 are those which are in the nullspace of the above matrix.
If we set x3=0, the non-trivial result eigenvector has x1=-x2... for simplicity, we say x1=1
==> x2=-1
hence the first eigenvector is (1,-1,0)
Next, set x3=1, we can now have x1=x2=0 without obtaining a trivial eigenvector, so the second eigenvector is (0,0,1) (although, you could also use non-zero x1 and x2, but this is the simplest case which preserves linear independence).
I hope this has helped :)