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?
Best Answer
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.
Have a go at proving the other set equality.