When there is a double root for the eigenvalue, how many eigenvectors?

[Petrus]

Hello MHB,
I got one question. If I want to find basis ker and it got double root in eigenvalue but in that eigenvalue i find one eigenvector(/basis) what kind of decission can I make? Is it that if a eigenvalue got double root Then it Will ALWAYS have Two eigenvector(/basis)?

Regards,
\(\displaystyle |\pi\rangle\)

 

[Opalg]

Re: 1 basis or Two basis for double root to ker?

If I want to find basis ker and it got double root in eigenvalue but in that eigenvalue i find one eigenvector(/basis) what kind of decission can I make? Is it that if a eigenvalue got double root Then it Will ALWAYS have Two eigenvector(/basis)?

Not necessarily. When there is a double root for the eigenvalue there will always be at least one eigenvector. There may or may not be a second, linearly independent, eigenvector. For example, the matrices $\begin{bmatrix}1&0\\ 0&1 \end{bmatrix}$ and $\begin{bmatrix}1&1\\ 0&1 \end{bmatrix}$ both have a repeated eigenvalue $1$, but the first one has two linearly independent eigenvectors and the second one only has one.