Question

To find the eigenvalues and eigenvectors, [\Ax=\lambda x\], must [\A\] always be an non-singular matrix? Can matrix [\A\] be any rectangular singular or square singular matrix to work?

Answer 1

To find the eigenvalues and eigenvectors,
\[Ax\]\[=lambda x\]

Must matrix A always be an non-singular matrix? Can matrix A be any rectangular singular or square singular matrix to work?

Answer 2

What do you mean by work? Are you asking the conditions such that A is diagonalizable?

Answer 3

I thinking to find the eigenvalues and eigenvectors of a matrix A, must matrix A have any condition to have eigenvalues and eigenvectors? Or is it any matrix, singular or non-singular, square or rectangle, have some eigenvalues and eigenvectors?

Answer 4

It does not depend on whether the matrix is singular (has a non-trivial nullspace) it depends on the characteristic polynomial.

Answer 5

Eigenvalue of 0 is non-invertible, I think.

Answer 6

Exactly, thats my point you can have singular matrices with eigenvalues

Answer 7

So I can say that any matrix has some form of eigenvalues(be it complex or real number) and eigenvectors?

Answer 8

Sure there will always be eigenvalues since the eigenvalues are the roots of the characteristic polynomial. Whether they are real or not depends on the characteristic polynomial.

Answer 9

Not all matrices have eigenvalues.

Answer 10

ohh...what kind of matrices may not have eigenvalues and eigenvectors?

Answer 11

The eigenvalues are the roots of the characteristic polynomial

Answer 12

If the roots are not real then you can say there are "no eigenvalues" as estudier stated.

Answer 13

eg (0,-1,1,0) half pi rotation no eigenvalues.

Answer 14

ohh...So if the eigenvalue turns out to be a complex form, it has "no eigenvalues". But it will still have a proper eigenvector right?

Answer 15

no

Answer 16

If it has an eigenvector it has a corresponding eigenvalue

Answer 17

Since its corresponding eigenvalue is a complex form. It can still have eigenvector too, isn't it?

Answer 18

Since \[Ax = \lambda x\]\[\text{If there is such an $x$ then there is a corresponding $\lambda$}\]

Answer 19

U can deal with a complex eigenvalue as well with a little more work.

Answer 20

If I have a complex number as my lambda value, I can have some eigenvector "x" too, isn't it? Maybe the eigenvector has complex number in it though. Is this right?

Answer 21

You have to decide whether your transformation is over the complex field or not.

Answer 22

If you are working with the reals then complex eigenvalues do not exist.

Answer 23

ohh... I see... Thanks for explaining it to me. Thanks a lot!! :)

Answer 24

One last thing, a matrix is considered diagonalizable if the dimension of each eigenspace is equal to the multiplicity of the corresponding eigenvalue.

Answer 25

If a matrix is diagonalizable you can form a basis, for the vector space, consisting of eigenvectors.

Answer 26

Thanks a lot, Alchemista! :)