Let there be $T:\mathbb{C}^8\rightarrow \mathbb{C}^8$

Such that $ T\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \\ x_{7} \\ x_{8} \end{array}\right)=\left(\begin{array}{c} 2\,x_{8}+x_{7}+3\,x_{6}+5\,x_{5}-x_{4}+3\,x_{3}+x_{2}+5\,x_{1} \\ 4\,x_{8}+5\,x_{7}+3\,x_{6}+4\,x_{5}-3\,x_{4}-2\,x_{3}+3\,x_{2}-3\,x_{1} \\ 5\,x_{8}+x_{7}-2\,x_{6}-3\,x_{5}+3\,x_{4}+5\,x_{3}-3\,x_{2}-3\,x_{1} \\ -2\,x_{8}+2\,x_{7}-x_{6}+2\,x_{5}-2\,x_{4}+5\,x_{3}+5\,x_{2}+x_{1} \\ -x_{8}-2\,x_{7}-3\,x_{6}+5\,x_{5}+x_{4}+4\,x_{3}-5\,x_{2}-3\,x_{1} \\ 5\,x_{8}-x_{7}+5\,x_{6}-3\,x_{5}-5\,x_{4}-3\,x_{3}-5\,x_{2}-3\,x_{1} \\ -3\,x_{8}+4\,x_{7}-4\,x_{6}+4\,x_{5}+5\,x_{4}+4\,x_{3}+2\,x_{2}-2\,x_{1} \\ -5\,x_{8}-2\,x_{7}+4\,x_{6}+x_{5}+4\,x_{4}-3\,x_{3}+5\,x_{2}-x_{1} \end{array}\right)$

Does $T$ has a Jordan normal form over $\mathbb{C}$?

Doesn't over $\mathbb{C}$ all matrices have a Jordan normal form?

  • 4
    $\begingroup$ the common term is "over" rather than "above". The difference is subtle, but "above" sounds wrong. $\endgroup$ – Omnomnomnom Sep 1 '15 at 21:15
  • $\begingroup$ @Omnomnomnom edited, sorry. $\endgroup$ – gbox Sep 1 '15 at 21:16
  • 5
    $\begingroup$ Since the characteristic polynomial of $T$ splits no matter what it is, it will have a Jordan canonical form. $\endgroup$ – walkar Sep 1 '15 at 21:38

Yes, all matrices over $\mathbb{C}$ have a Jordan normal form.

Furthermore, if $A$ is a square matrix with entries in a field $F$ and $F$ is an algebraically closed field (so the characteristic equation $c_A(x)$ splits over $F$), then $A$ has a Jordan normal form.

If $F$ is not algebraically closed, then there exists a field extension $K$ > $F$ such that $c_A(x)$ does split over $K$.


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