# Diagonalizable Linear Operator imples $V=\operatorname{Ker}(T) \oplus \operatorname{Im}(T)$

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?

Every $$v \in V$$ can be (uniquely) expressed as $$v = a_1 v_1 + \ldots + a_n v_n$$ for some scalars $$a_1, \ldots, a_n$$. Taking $$T$$ of both sides: \begin{align*} T(v) &= a_1 T(v_1) + \ldots + a_k T(v_k) + a_{k+1} T(v_{k+1}) + \ldots + a_n T(v_n) \\ &= a_1 \lambda_1 v_1 + \ldots + a_k \lambda_k v_k + 0 + \ldots + 0 \\ &= a_1 \lambda_1 v_1 + \ldots + a_k \lambda_k v_k. \end{align*} Now, if $$v \in \operatorname{Ker} T$$, then $$a_1 \lambda_1 v_1 + \ldots + a_k \lambda_k v_k = 0.$$ Recall that $$(v_1, \ldots, v_k)$$ is linearly independent, so we can conclude that, $$a_1 \lambda_1 = \ldots = a_k \lambda_k = 0,$$ and since $$\lambda_i \neq 0$$ for $$i = 1, \ldots, k$$, we have $$a_1 = \ldots = a_k = 0.$$ Thus, $$v = 0v_1 + \ldots + 0v_k + a_{k+1}v_{k+1} + \ldots + a_n v_n \in \operatorname{span}(v_{k+1}, \ldots, v_n),$$ hence $$\operatorname{Ker} T \subseteq \operatorname{span}(v_{k+1}, \ldots, v_n).$$ Your observation proves the other inclusion.