Does an eigenbasis of $R^n$ have to consist of vectors corresponding to different eigenvalues?

Question

Does an Eigenbasis of $R^n$ have to consist of vectors corresponding to different eigenvalues? I am reading up on diagonalization, but this detail is not clear to me.

Answer 1

It depends on what you mean by that question.

For each eigenvalue, you will have to have a number of linearly independent eigenvectors for it equal to the multiplicity of the eigenvalue (i.e., the number of times the eigenvalue is a root of the characteristic equation). So you will need eigenvectors for each eigenvalue, but each eigenvalue can have more than one basis eigenvector.

In Cameron Williams example, the characteristic equation is $(x - 1)^n = 0$, so $1$ is an eigenvector of multiplicity $n$, and therefore you need $n$ eigenvectors of it.

For the matrix $$\begin{bmatrix} 1 & 0 & 0\\0& 1&0\\0&0&2\end{bmatrix}$$ $(1,0,0)$ and $(0,1,0)$ are eigenvectors of $1$, while $(0,0,1)$ is an eigenvector of $2$.