Is it possible to Diagonalize a matrix with repeated eigenvalues?

Is it possible to Diagonalize a matrix with repeated eigenvalues?

A matrix with repeated eigenvalues can be diagonalized. Just think of the identity matrix. All of its eigenvalues are equal to one, yet there exists a basis (any basis) in which it is expressed as a diagonal matrix.

Can repeated eigenvalues have different eigenvectors?

When eigenvalues of the matrix A are repeated with a multiplicity of r, some of the eigenvectors may be linearly dependent on others. Guidance as to the number of linearly independent eigenvectors can be obtained from the rank of the matrix A.

Are eigenvectors of repeated eigenvalues linearly independent?

Eigenvectors corresponding to distinct eigenvalues are always linearly independent. It follows from this that we can always diagonalize an n × n matrix with n distinct eigenvalues since it will possess n linearly independent eigenvectors.

What if all the eigen values are same?

In particular, what happens if all eigenvalues are all equal to 1? An n×n matrix with an eigenvalue 1 of multiplicity n is called a unipotent matrix, while a matrix with a full set of identical eigenvalues is said to be projectively unipotent.

Do same eigenvalues have same eigenvectors?

The set of all eigenvectors of T corresponding to the same eigenvalue, together with the zero vector, is called an eigenspace, or the characteristic space of T associated with that eigenvalue. If a set of eigenvectors of T forms a basis of the domain of T, then this basis is called an eigenbasis.

Are repeated eigenvalues stable?

If the two repeated eigenvalues are positive, then the fixed point is an unstable source. If the two repeated eigenvalues are negative, then the fixed point is a stable sink.

Can an eigenvalue be repeated?

We say an eigenvalue λ1 of A is repeated if it is a multiple root of the characteristic equation of A—in other words, the characteristic polynomial |A − λI| has (λ − λ1)2 as a factor. Let’s suppose that λ1 is a double root; then we need to find two linearly independent solutions to the system (4) corresponding to λ1.