Number of linear independent eigenvectors of a matrix

Question

Consider matrix A n*n with rank r and m distinct eigenvalues. I know that there is an independent eigenvector corresponds to each eigenvalues. Is it possible to have more independent eigenvectors than distinct eigenvalues? In general case is there any relation between n , r and m and number of independent eigenvectors?

Answer 1

The only general result is this: the eigenvalues of an $n\times n$ matrix $A$ are the roots of its characteristic polynomial $\det(A-tI)$, which has degree $n$, so there are $m\le n$ distinct real or complex eigenvalues $\lambda_1,\dots,\lambda_m$, each with algebraic multiplicity $r_i$, and of course $$r_1+\dots+r_m=n.$$ On the other hand, to each eigenvalue $\lambda_i$ there corresponds an eigenspace $E_{\lambda_i}$. The dimensions of eigenspaces are called the geometric multiplicities of the eigenvalues, and they satisfy the inequalities $$1\le \dim E_{\lambda_i}\le r_i\qquad(i=1,\dots,m).$$ A necessary and sufficient condition for a matrix to be diagonalisable is that the geometric multiplicity of each eigenvalue be equal to its algebraic multiplicity.