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Prove that the matrix

$$A = \begin{bmatrix} 0 & 1 & \\ \ddots & & \ddots \\ & \ddots & & \ddots \\ & & \ddots & & \ddots\\ & & & 0 & & 1\\ -a_{0} & -a_{1} & \dots & \dots & \dots & -a_{n-1}\end{bmatrix}$$

is non-derogatory. I am not sure how would I do this. Non-derogatory matrix means that the minimal polynomial equals the characteristic polynomial but I am not sure how to this one.

Rodrigo de Azevedo
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Fernando Martinez
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    See Companion matrix. – Git Gud Nov 13 '16 at 21:17
  • Yes a companion matrix the charectertistic polynomial equals the minimal polynomial – Fernando Martinez Nov 13 '16 at 21:22
  • So wait would would this be the transpose of the comparison matrix – Fernando Martinez Nov 13 '16 at 21:57
  • My proof is the following the transpose of my matrix $A$ is $A^T$ and the tranpose of A is the companion matrix. And A is always similar to its tranpose. And if two matrices are similar their characteristic polynomial and the minimum polynomial so then let P be the charecteristic polynomial and Q be the mimum polynomial so $P_{A}(t)=P_{A^T}(T)=Q_{A^T}(t)=Q_{A^T}(t)$ thus a is non derogatory – Fernando Martinez Nov 14 '16 at 03:09
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    @FernandoMartinez: While all that is true, it uses the somewhat non-obvious (an unnatural) fact that a matrix is always similar to its transpose. It is easier to see that minimal and characteristic polynomials of $A^T$ are always the same as the respective ones for $A$, which suffices here to deduce the result from the one for companion matrices. – Marc van Leeuwen Nov 14 '16 at 16:24

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