# Diagonalizability condition involving direct sum of nullspace and range

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.