Let $T$ be a linear operator on a finite-dimensional space $V$. Let $c_l,…,c_k$ be the distinct characteristic values of $T$ and let $W_i$ be the null space of $(T-c_iI)$. The following are equivalent.
(1) $T$ is diagonalizable.
(2) The characteristic polynomial for $T$ is $f=(x-c_1)^{d_1}\dotsb (x-c_k)^{d_k}$ and $\text{dim}(W_i)=d_i$, $i=1,…,k$.
(3) $\text{dim}(W_1)+…+\text{dim}(W_k)=\text{dim}(V)$.
Now there are lots of way to prove this theorem, like $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$, I think this is the least number of implication needed to prove this theorem and $(1)\Leftrightarrow (2)$, $(1) \Leftrightarrow (3)$, $(2)\Leftrightarrow (3)$ this is maybe the most number of implication needed to prove this theorem. I want to show $(1)\Leftrightarrow (2)\Rightarrow (3)\Rightarrow (1)$.
My attempt: Here is proof of $(1)\Rightarrow (2)$. For more detail proof see comment.
$(2)\Rightarrow (3)$. Suppose characteristic polynomial of $T$ is $f=(x-c_1)^{d_1}\dotsb (x-c_k)^{d_k}$ and $\text{dim}(W_i)=d_i$, $\forall i\in J_k$. It’s standard result, $f(x)=(-1)^{n}x^n+(-1)^{n-1}\text{tr}(T)x^{n-1}+…+\text{det}(T)$, $\forall x\in F$. Since $(-1)^n\neq 0$, we have $\text{deg}(f)=n=\dim(V)$. By theorem 1 section 4.2, $\deg(f)$ $=\text{deg}[(x-c_1)^{d_1}\dotsb (x-c_k)^{d_k}]$ $=\deg [(x-c_1)^{d_1}]+…+\deg[(x-c_k)^{d_k}$ $=d_1+…+d_k$. Hence $\dim(V)=d_1+…+d_k=\dim(W_1)+…+\dim(W_k)$.
$(3)\Rightarrow (1)$. Suppose $\text{dim}(W_1)+…+\text{dim}(W_k)=\text{dim}(V)$. Let $W=W_1+…+W_k$. By lemma 3 section 6.2, $\dim(W)= \text{dim}(W_1)+…+\text{dim}(W_k)=\text{dim}(V)$. So $W=V$. Let $B_i$ be basis of $W_i$. By lemma 3 section 6.2, $B=\bigcup_{i=1}^k B_i$ is basis of $V$. Thus $\exists B$ basis of $V$ such that $\forall \alpha \in B$, $\alpha$ is eigenvector of $T$. Hence $T$ is diagonalizable.
How to show $(2)\Rightarrow (1)$?
Edit: Proof of $(2)\Rightarrow (1)$ is generalization of this proof of mine. Which is essentially $(2)\Rightarrow (3)\Rightarrow (1)$.
