Can you have two of the same eigenvectors?

Can you have two of the same eigenvectors?

Matrices can have more than one eigenvector sharing the same eigenvalue. The converse statement, that an eigenvector can have more than one eigenvalue, is not true, which you can see from the definition of an eigenvector.

Do matrices with same eigenvalues have same eigenvectors?

If two matrices are similar, they have the same eigenvalues and the same number of independent eigenvectors (but probably not the same eigenvectors). If two matrices have the same n distinct eigenvalues, they’ll be similar to the same diagonal matrix.

Can 2 eigenvectors have the same eigenvalues?

Two similar matrices have the same eigenvalues, even though they will usually have different eigenvectors. Said more precisely, if B = Ai’AJ. I and x is an eigenvector of A, then M’x is an eigenvector of B = M’AM. So, A1’x is an eigenvector for B, with eigenvalue ).

Can you have two of the same eigenvalues?

Two distinct eigenvalues for the same would mean is somehow stretched by two different amounts. For example, if the eigenvalues were 2 and 3, the scale of the axis would be both doubled and tripled. The only way this could possibly make sense is if is 0, but eigenvectors are non-zero by definition. So it can’t be.

Can a 3×3 matrix have 4 eigenvectors?

So it’s not possible for a 3 x 3 matrix to have four eigenvalues, right? right.

Do similar matrices have the same eigenvalues?

Since similar matrices A and B have the same characteristic polynomial, they also have the same eigenvalues. Thus Pv (which is non-zero since P is invertible) is an eigenvector for B with eigenvalue λ.

Can a 3×3 matrix have more than 3 eigenvectors?

To be pedantic, the “number of eigenvectors” of a matrix is always infinite when working with the real or complex numbers (or any other infinite field), since if are eigenvectors for an eigenvalue , then so is every non-zero vector in .

How many eigenvectors can a matrix have?

EDIT: Of course every matrix with at least one eigenvalue λ has infinitely many eigenvectors (as pointed out in the comments), since the eigenspace corresponding to λ is at least one-dimensional.