Let $T$ be a linear operator on the $n$-dimensional vector space $V$, and suppose that $T$ has $n$ distinct characteristic values. Prove that $T$ is diagonalizable.
We will use following lemma: (1) Let $\lambda_1,…,\lambda_m$ ($m ≤ n$) be the distinct eigenvalues of $T$ and $v_1,…,v_m$ be corresponding eigenvectors. Then $v_1,…,v_m$ are linearly independent. (2) Let $\lambda$ be an eigenvalue of $T$. The geometric multiplicity of $\lambda$ is less than or equal to the algebraic multiplicity of $\lambda$.
Approach (1): Let $\lambda_1,…,\lambda_n$ be eigenvalue of $T$. So $\forall i\in J_n$, $\exists v_i\in V\setminus \{0_V\}$ such that $T(v_i)=\lambda_i\cdot v_i$. By lemma (1), $\{v_1,…,v_n\}$ is independent. Since $\dim (V)=n$, we have $\{v_1,…,v_n\}$ is basis of $V$. Thus $\exists B=\{v_1,…,v_n\}$ basis of $V$ such that $\forall i\in J_n$, $v_i$ is eigenvector of $T$. Hence $T$ is diagonalizable.
Approach (2): Let $\lambda_1,…,\lambda_n$ be eigenvalue of $T$. Then characteristic polynomial of $T$ is $f:F\to F$ such that $f(x)=(x-\lambda_1)\dotsb (x-\lambda_n)$, $\forall x\in F$. Let $W_i$ be eigenspace of $\lambda_i$, i.e. $W_i=N_{T-\lambda_i I}$. By lemma (2), $0\leq \dim(W_i)\leq$ algebraic multiplicity of $\lambda_i$ $=1$. Since $\exists 0_V\neq v_i\in W_i$, we have $\dim (W_i)\neq 0$. So $\dim (W_i)=1$, $\forall i\in J_n$. By theorem 2 section 6.2, $T$ is diagonalizable. Is my proof correct?
