I'm working on the following problem from Axler's Linear Algebra Done Right (Exercise 5.C.5):

Suppose $V$ is a finite-dimensional complex vector space and $T\in \cal{L}(V)$. Prove that $T$ is diagonalizable if and only if $$V=\text{null}(T-\lambda I)\oplus\text{range}(T-\lambda I)$$ for every $\lambda\in \mathbb{C}$.

I've already proven the "only if" direction (sketch: suppose $V$ has a basis of eigenvectors of $T$, show that $V=\text{null}(T-\lambda I)+\text{range}(T-\lambda I)$, and conclude using the rank-nullity theorem that the sum is actually a direct sum), but I'm having trouble proving the other direction.

My attempt: Let $\lambda_1,\ldots,\lambda_m$ denote the distinct eigenvalues of $T$. We know that $V=\text{null}(T-\lambda_i I)\oplus\text{range}(T-\lambda_i I)$ for every $i$, and we want to show that (for example) $V=\text{null}(T-\lambda_1 I)+\cdots+\text{null}(T-\lambda_m I)$.

Any hints would be appreciated.



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