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$F$ ia a field. a matrix $A\in M_n(F)$ is called cyclic if there is a vector $v\in M_{1\times n}(F)$ such that $\{v, vA, . . . , vA^{n-1}\}$ is a basis for $M_{1\times n}(F)$ as a left vector space over $F$. I have to show , the representation of A in the above basis has the following form ‎$\begin{pmatrix}‎ ‎0 & 1 & 0 & \ldots & 0 \\ 0 & 0 &1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 \ldots & \ldots & 0 & 1 \\ a_{1} & a_{2} & \ldots & a_{n-1} & a_{n} ‎ ‎\end{pmatrix}$‎, for som $a_1‎ , ‎a_2 ,‎ ‎\ldots‎ , ‎a_n \in F$‎. ‎

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marked as duplicate by Travis, N. F. Taussig, graydad, user85798, Empty Oct 31 '15 at 2:58

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Hint: There exists $a_1,...,a_n\in F$ such that $$vA^n=a_1v+a_2vA+...+a_nvA^{n-1}$$ and for every $i<n-1$, $(vA^i)A=vA^{i+1}$.

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  • $\begingroup$ Hi, Thanke you Nitrogen. Can I ask you explain more, please. By you're answer the representation of A is $\begin{pmatrix}‎ ‎0 & 0 &\ldots & 0 &‎ -‎a_1\\‎ ‎1 &0 & \ldots & 0 &‎ -a_2\\‎ ‎0 & 1 & \ldots & 0 &‎ -‎a_3\\‎ ‎\vdots & \vdots & \ddots & \vdots & \vdots\\‎ ‎0 & 0 & \ldots & 1 &‎ -‎a_{n}‎ ‎\end{pmatrix}.$ not above? $\endgroup$ – math123 Oct 30 '15 at 19:41
  • $\begingroup$ Yes exactly, but now you have to be careful because you wrote you question with multiplication on the right by $A$, so you should take the transpose of this matrix, and the result will be in the desired form. $\endgroup$ – Nitrogen Oct 30 '15 at 19:43
  • $\begingroup$ Ok, multiplication on the right by $A$ But Why we take the transpose of this matrix?Cen you tell me please. $\endgroup$ – math123 Oct 30 '15 at 19:49
  • $\begingroup$ Because $(vA)^t=A^tv^t$. $\endgroup$ – Nitrogen Oct 30 '15 at 19:50
  • $\begingroup$ yes. Thanke you Nitrogen, thanke you very much. $\endgroup$ – math123 Oct 30 '15 at 19:51

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