# About test for diagonalizability. Question about $\operatorname{rank}(T-\lambda I)$

"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?

• Use \operatorname{rank} – Invisible Mar 3 at 10:42

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.