Suppose that $T\in L(V)$ is diagonalizable. Prove that $V=\operatorname{Ker}(T) \oplus \operatorname{Im}(T)$
A solution I found online follows: Since $T$ is diagonalizable there exits a basis for $V$ consisting of eigenvectors of $T$. Call this basis $v_1,v_2,\ldots,v_n$ then we split the basis into vectors which correspond to nonzero eigenvalues and those which correspond to zero eigenvalues. Say that $\lambda_1,\lambda_2,\ldots,\lambda_k$ are nonzero eigenvalues corresponding to $v_1,v_2,\ldots,v_k$ then $v_{k+1},\ldots,v_n$ correspond to $\lambda=0$.
Then they claim that, $$\operatorname{Ker}(T)=\operatorname{span}(v_{k+1},\ldots,v_n)$$ $$\operatorname{Im}(T)=\operatorname{span}(v_1,\ldots,v_k)$$ Why is this true? Is it simply because $\operatorname{span}(v_{k+1},\ldots,v_n) = c_{k+1} \lambda_{k+1} v_{k+1} + \cdots + c_n \lambda_n v_n = 0$ for all values of $c_i$ ? If the case for the Image is just that all nonzero vectors can be written as a linear combination of the basis vectors why do we require that these correpond to nonzero eigenvalues?