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"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?

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    $\begingroup$ Use \operatorname{rank} $\endgroup$ – Invisible Mar 3 at 10:42
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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.

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