Let $T: V \to V$ be a linear operator $V$ and a vector space over $\mathbb K$.If $p_T$ has all its roots in $\mathbb K$ and if they are simple (i.e with algebraic multiplicity 1), then $T$ is diagonalizable.
I found this statement (no demo) and I didn't get it.
If the algebraic multiplicities are 1 means that all eigenvalues of A are different but that means linearly independent eigenvectors (basis of eigenvectors) so A is diagonalizable?
Yes, if the characteristic polynomial of an $n$ x $n$ matrix $T$ has $n$ distinct roots, then there exists a matrix $A$ of eigenvalues $\lambda_i$ of $T$ and a change-of-basis matrix $S$ such that $T = S^{-1}AS$. Suppose the resulting eigenvectors $\{x_{\lambda_i}\}$ were linearly dependent. Then without loss of generality we have $x_{\lambda_1} = \sum_{i=2}^{n} \alpha_i x_{\lambda_i} \Rightarrow Tx_{\lambda_1} = \lambda_1x_{\lambda_1} = T\sum_{i=2}^{n} \alpha_i x_{\lambda_i} = \sum_{i=2}^{n} \alpha_i Tx_{\lambda_i} =\sum_{i=2}^{n} \alpha_i \lambda_ix_{\lambda_i}$.
However, since each $\lambda_i$ is distinct, $\lambda_1 \sum_{i=2}^{n} \alpha_i x_{\lambda_i} \neq \sum_{i=2}^{n} \alpha_i \lambda_ix_{\lambda_i}$, a contradiction.
Your proof is kind of correct. “$p_T$ has all its roots in $\mathbb K$ and if they are simple (i.e with algebraic multiplicity 1)” means $p_T(x)=(x-\lambda_1)\dotsb (x-\lambda_n)$, $\forall x\in \Bbb{K}$. Let $W_i$ be eigenspace of $\lambda_i$, i.e. $W_i=N_{T-\lambda_i I}$. It’s easy to check, $0\leq \dim(W_i)\leq$ algebraic multiplicity of $\lambda_i$ $=1$. Since $\exists 0_V\neq v_i\in W_i$, we have $\dim (W_i)\neq 0$. So $\dim (W_i)=1$, $\forall i\in J_n$. let $B_i=\{\alpha_i\}$ be basis of $W_i$. We claim $B=\bigcup_{i=1}^n B_i=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. If $c_1\cdot \alpha_1+…+c_n\cdot \alpha_n=0$. Since $W_1\oplus \dotsb \oplus W_n$, we have $c_i\cdot \alpha_i=0$. So $c_i=0$, $\forall i\in J_n$. Thus $\{\alpha_1,…,\alpha_n\}$ is independent. Since $\dim(V)=n$, we have $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. Hence $\exists B$ basis of $V$ such that $\forall i\in J_n$, $\alpha_i$ is eigenvector of $T$. So $T$ is diagonalizable.
Alternative proof: Suppose $\lambda_1,…,\lambda_n$ are distinct eigenvalues of $T$. Then $\forall i\in J_n$, $\exists 0_V\neq\alpha_i\in V$ such that $T(\alpha_i)=\lambda_i\cdot \alpha_i$. By this post, $\{\alpha_1,…,\alpha_n\}$ is independent. Since $\dim(V)=n$, we have $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. Hence $\exists B$ basis of $V$ such that $\forall i\in J_n$, $\alpha_i$ is eigenvector of $T$. So $T$ is diagonalizable.
