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Homeproof of necessary and sufficient condition for diagonalizability

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# 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 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 $v$, $Z(v,T)=\left<v,Tv,\ldots,T^{{r}}v\right>$. 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.

## Mathematics Subject Classification

15A04*no label found*

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