Eigenvectors and Eigenvalues

  • Thread starter DiamondV
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  • #1
DiamondV
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Homework Statement


Find the eigenvalues and associated eigenvector of the following matrix:
32e0fe9a83.png


Homework Equations




The Attempt at a Solution


302cba53a4.jpg


We have a theorem in our lectures notes that states that if a matrix is invertible the only eigenvector in its kernel will be the zero vector. In order to find out if it is invertible we get the det(A) and see if its equal to 0 or not, if it is equal to 0(you can't divide by 0) then there is no inverse, if it is not equal to 0(like in this case I got 6) then it is invertible and the only vector is the zero vector in the kernel. So technically I should stop my calculations at this point and say the zero vector is the only one.
However in the solutions given to use they have an answer that is not a 0 vector.
3b0554f97a.png

1,2,3 are eigenvalues.
How is this possible?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Find the eigenvalues and associated eigenvector of the following matrix:
32e0fe9a83.png


Homework Equations




The Attempt at a Solution


302cba53a4.jpg


We have a theorem in our lectures notes that states that if a matrix is invertible the only eigenvector in its kernel will be the zero vector. In order to find out if it is invertible we get the det(A) and see if its equal to 0 or not, if it is equal to 0(you can't divide by 0) then there is no inverse, if it is not equal to 0(like in this case I got 6) then it is invertible and the only vector is the zero vector in the kernel. So technically I should stop my calculations at this point and say the zero vector is the only one.
However in the solutions given to use they have an answer that is not a 0 vector.
3b0554f97a.png

1,2,3 are eigenvalues.
How is this possible?

There is no such theorem as the one you state from your lectures. The zero vector is not considered as an eigenvector at all.

There is a theorem stating that if a matrix is invertible (has non-zero determinant) then the only vector (NOT EIGENVECTOR!) in the kernel is the zero vector. That has nothing at all to do with eigenvalues and eigenvectors.

Do you actually know what eigenvalues and eigenvectors are?
 
  • #3
DiamondV
103
0
There is no such theorem as the one you state from your lectures. The zero vector is not considered as an eigenvector at all.

There is a theorem stating that if a matrix is invertible (has non-zero determinant) then the only vector (NOT EIGENVECTOR!) in the kernel is the zero vector. That has nothing at all to do with eigenvalues and eigenvectors.

Do you actually know what eigenvalues and eigenvectors are?

Oh. Not really. Our lecture notes haven't shown any graphs with vectors on them or any sort of visualisation for this. I just know that there's these things called eigenvectors and eigenvalues that are really useful for some reason.
 

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