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Suppose that $A=C\oplus C=\begin{pmatrix} C & 0 \\ 0& C \end{pmatrix}$, $C$ be a companion matrix of $m(x)=m_0+m_1x+\ldots+m_{n-1}x^{n-1}+x^n$ . I have to show that $A$ is not cyclic. Can any one help?

Harish Chandra Rajpoot
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q1000
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    Could you explain the definition of "$A$ is cyclic" – quid Jul 13 '15 at 11:33
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    a matrix $A\in Mn(F)$ is called cyclic if there is a vector $t \in M_{n\times 1}(F)$ such that ${\alpha, \alpha A, . . . , \alpha A^{n−1}}$ is a basis for $M_{1×n}(F)$ as a left vector space over $F$. – q1000 Jul 13 '15 at 11:38

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A space is cyclic with respect to a linear transformation $T$ iff the minimal polynomial of $T$ coincides with its characteristic polynomial.

The minimal polynomial of $A$ is $m$ but its characteristic polynomial is $m^2$. Since they are different, $A$ is not cyclic.

lhf
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  • See http://math.stackexchange.com/questions/1035393/minimal-polynomials-and-cyclic-subspaces. – lhf Jul 13 '15 at 11:41
  • thanke you lhf. Can you explain more why the minimal polynomial of A is m, please. – q1000 Jul 13 '15 at 13:24
  • @q1000, $m$ is the minimal polynomial of $C$ and so $m(A)=0$. You just need this to know that the minimal polynomial of $A$ is different from its characteristic polynomial. – lhf Jul 13 '15 at 13:33