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Homeproof of Jordan canonical form theorem

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# proof of Jordan canonical form theorem

This theorem can be proved combining the cyclic decomposition theorem and the primary decomposition theorem. By hypothesis, the characteristic polynomial of $T$ factorizes completely over $F$, and then so does the minimal polynomial of $T$ (or its annihilator polynomial). This is because the minimal polynomial of $T$ has exactly the same factors on $F[X]$ as the characteristic polynomial of $T$. Let’s suppose then that the minimal polynomial of $T$ factorizes as $m_{{T}}=(X-\lambda_{1})^{{\alpha_{1}}}\ldots(X-\lambda_{r})^{{\alpha_{r}}}$. We know, by the primary decomposition theorem, that

$V=\bigoplus_{{i=1}}^{{r}}\ker((T-\lambda_{i}I)^{{\alpha_{i}}}).$ |

Let $T_{{i}}$ be the restriction of $T$ to $\ker((T-\lambda_{i}I)^{{\alpha_{{i}}}})$. We apply now the cyclic decomposition theorem to every linear operator

$(T_{{i}}-\lambda_{i}I)\colon\ker(T-\lambda_{i}I)^{{\alpha_{{i}}}}\to\ker(T-% \lambda_{i}I)^{{\alpha_{{i}}}}.$ |

We know then that $\ker(T-\lambda_{i}I)^{{\alpha_{i}}}$ has a basis $B_{{i}}$ of the form $B_{{i}}=B_{{1,i}}\bigcup B_{{2,i}}\bigcup\ldots\bigcup B_{{d_{i},i}}$ such that each $B_{{s,i}}$ is of the form

$B_{{s,i}}=\{v_{{s,i}},(T-\lambda_{i})v_{{s,i}},(T-\lambda_{i})^{{2}}v_{{s,i}},% \ldots,(T-\lambda_{i})^{{k_{{s,i}}}}v_{{s,i}}\}.$ |

Let’s see that $T$ in each of this “cyclic sub-basis” $B_{{s,i}}$ is a Jordan block: Simply notice the following fact about this polynomials:

$\displaystyle X(X-\lambda_{i})^{{j}}$ | $\displaystyle=$ | $\displaystyle(X-\lambda_{i})^{{j+1}}+X(X-\lambda_{i})^{{j}}-(X-\lambda_{i})^{{% j+1}}$ | ||

$\displaystyle=$ | $\displaystyle(X-\lambda_{i})^{{j+1}}+(X-X+\lambda_{i})(X-\lambda_{i})^{{j}}$ | |||

$\displaystyle=$ | $\displaystyle(X-\lambda_{i})^{{j+1}}+\lambda_{i}(X-\lambda_{i})^{{j}}$ |

and then

$T(T-\lambda_{i}I)^{{j}}(v_{{s,i}})=(T-\lambda_{i})^{{j+1}}(v_{{s,i}})+\lambda_% {i}(T-\lambda_{i}I)^{{j}}(v_{{s,i}}).$ |

So, if we also notice that $(T-\lambda_{i}I)^{{k_{{s,i}}+1}}(v_{{s,i}})=0$, we have that $T$ in this sub-basis is the Jordan block

$\begin{pmatrix}\lambda_{i}&0&0&\cdots&0&0\\ 1&\lambda_{i}&0&\cdots&0&0\\ 0&1&\lambda_{i}&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&\lambda_{i}&0\\ 0&0&0&\cdots&1&\lambda_{{i}}\end{pmatrix}$ |

So, taking the basis $B=B_{{1}}\bigcup B_{{2}}\bigcup\ldots\bigcup B_{{r}}$, we have that $T$ in this basis has a Jordan form.

This form is unique (except for the order of the blocks) due to the uniqueness of the cyclic decomposition.

## Mathematics Subject Classification

15A18*no label found*

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