"Let T be a linear operator on an n-dimensional vector space V. Then T is diagonalizaable if and only if:
1).The characteristic polynomial T splits.
2). For each eigenvalue $\lambda$ of T, the multiplicity of $\lambda$ equals $n-\operatorname{rank}(T-\lambda I)$
Question: Is $\operatorname{rank}(T-\lambda I)$ the dimension of the eigenspace which makes up the eigenvectors? Why does $n-\operatorname{rank}(T-\lambda I)$ have to equal the multiplicity?
Best Answer
Represent $T$ by a matrix $A.$ Then $T- \lambda I$ is represented by $A- \lambda I$. For any $m \times n$ matrix $B$ the dimensiion of the solution space of $B$ is $n-\text {rank} (B)$. So all that the second condition is saying is that the algebraic multiplicity of the eigenvalue $\lambda$ equals the dimension of the corresponding eigenspace.