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# Finding eigenvectors of matrix watch

1. Hello, I'm trying to find the eigenvectors of the matrix above. From solving the characteristic polynomial, I get the eigenvalues as 1, 1 and 2.

For the eigenvalue of 1, I then get a resulting equation x + 2y + 3z = 0

I could set:

x=-2, y=1, z=0 --> (-2, 1, 0)
x=-3, y=0, z=1 --> (-3, 0, 1)
x=0, y=3, z=-2 --> (0, 3, -2)

How do I know which are the two correct eigenvectors?
2. (Original post by Serendreamers)

Hello, I'm trying to find the eigenvectors of the matrix above. From solving the characteristic polynomial, I get the eigenvalues as 1, 1 and 2.

For the eigenvalue of 1, I then get a resulting equation x + 2y + 3z = 0

I could set:

x=-2, y=1, z=0 --> (-2, 1, 0)
x=-3, y=0, z=1 --> (-3, 0, 1)
x=0, y=3, z=-2 --> (0, 3, -2)

How do I know which are the two correct eigenvectors?
I've not checked your working, but those three vectors are linearly dependent so any two would suffice.
3. (Original post by Serendreamers)

Hello, I'm trying to find the eigenvectors of the matrix above. From solving the characteristic polynomial, I get the eigenvalues as 1, 1 and 2.

For the eigenvalue of 1, I then get a resulting equation x + 2y + 3z = 0

I could set:

x=-2, y=1, z=0 --> (-2, 1, 0)
x=-3, y=0, z=1 --> (-3, 0, 1)
x=0, y=3, z=-2 --> (0, 3, -2)

How do I know which are the two correct eigenvectors?
I am no expert in linear algebra but if you assume λ=1

then
-2x-2y+0z=1x
x+3y+3z=1y
0x+0y+z=1z

first equation gives 2y = -3x
second equation now becomes 2x = 3z

set x=6

gives (6,-9,4)

PS I hope I have no numerical mistakes because I do not have pen and paper in front of me.

EDIT Ignore all wrong I found some paper so I will recalculate
4. (Original post by joostan)
I've not checked your working, but those three vectors are linearly dependent so any two would suffice.
Ok, that makes sense.

I have another question though. If I ended up with x=0, y=0 and z=0, would my resulting eigenvector be (0, 0, 0) or (1, 1, 1)?
5. I am no expert in linear algebra but if you assume λ=1

then
0x-2y-2z=1x
x+3y+3z=1y
0x+0y+z=1z

first equation gives x+2y +2z=0
second equation gives 2x +2y+3z= 0
implication z=0
x=-2y

gives (-2,1,0)
6. (Original post by Serendreamers)
Ok, that makes sense.

I have another question though. If I ended up with x=0, y=0 and z=0, would my resulting eigenvector be (0, 0, 0) or (1, 1, 1)?
I guess (0, 0, 0) is like an eigenvector in the sense that it has no direction, and that remains unchanged and is always a trivial solution to:
but you're not really interested in that.

(Original post by TeeEm)
I am no expert in linear algebra but if you assume λ=1

then
0x-2y-2z=1x
x+3y+3z=1y
0x+0y+z=1z

first equation gives x+2y +2z=0
second equation gives 2x +2y+3z= 0
implication z=0
x=-2y

gives (-2,1,0)

Are you sure? Subbing in into the matrix seems to give the equation the OP gave. . .
7. (Original post by joostan)
I guess (0, 0, 0) is always an eigenvector in the sense that it has no direction
No. the zero vector is NEVER an eigenvector (by definition)

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