proof of necessary and sufficient condition for diagonalizability

First, suppose that $T$ is diagonalizable. Then $V$ has a basis whose elements $\{v_{1},\ldots,v_{n}\}$ are eigenvectors of $T$ associated with the eigenvalues $\{\lambda_{1},\ldots,\lambda_{n}\}$ respectively. For each $i=1,...,n$, as $v_{i}$ is an eigenvector, its annihilator polynomial is $m_{v_{i}}=X-\lambda_{i}$. As these vectors form a basis of $V$, we have that the minimal polynomial (http://planetmath.org/MinimalPolynomialEndomorphism) of $T$ is $m_{T}=\operatorname{lcm}(X-\lambda_{1},\ldots,X-\lambda_{n})$ which is trivially a product of linear factors.

Now, suppose that $m_{T}=(X-\lambda_{1})\ldots(X-\lambda_{p})$ for some $p$. Let $v\in V$. Consider the $T$ - cyclic subspace generated by $v$, $Z(v,T)=\left$. Let $T_{v}$ be the restriction of $T$ to $Z(v,T)$. Of course, $v$ is a cyclic vector of $Z(v,T_{v})$, and then $m_{v}=m_{T_{v}}=\chi_{T}$. This is really easy to see: the dimension of $Z(v,T)$ is $r+1$, and it’s also the degree of $m_{v}$. But as $m_{v}$ divides $m_{T_{v}}$ (because $m_{T_{v}}v=0$), and $m_{T}$ divides $\chi_{T_{v}}$ (Cayley-Hamilton theorem), we have that $m_{v}$ divides $\chi_{T_{v}}$. As these are two monic polynomials of degree $r+1$ and one divides the other, they are equal. And then by the same reasoning $m_{v}=m_{T_{v}}=\chi_{T}$. But as $m_{v}$ divides $m_{T}$, then as $m_{v}=m_{T_{v}}$, we have that $m_{T_{v}}$ divides $m_{T}$, and then $m_{T_{v}}$ has no multiple roots and they all lie in $k$. But then so does $\chi_{T_{v}}$. Suppose that these roots are $\lambda_{1},\ldots,\lambda_{r+1}$. Then $Z(v,T)=\bigoplus_{\lambda_{i}}E_{\lambda_{i}}$, where $E_{\lambda_{i}}$ is the eigenspace associated to $\lambda_{i}$. Then $v$ is a sum of eigenvectors. QED.

Title proof of necessary and sufficient condition for diagonalizability ProofOfNecessaryAndSufficientConditionForDiagonalizability 2013-03-22 14:15:45 2013-03-22 14:15:45 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Proof msc 15A04