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Let $A$ be a $n \times n$ matrix over the field $\mathbb C$ and $X \in \mathbb C^{n}$. Let $k$ be the least positive integer such that the set of vectors $\{X, AX, A^{2} X, \dots , A^{k} X \}$ are lineary dependent.

Now let us consider a relation $\sum_{i=1}^{k} c_{i} A^{i} X = 0$ and a polynomial $g(t) = \sum_{i=1}^{k} c_{i} t^{i}$. Then I have proved that each root of the equation $g(t) = 0$ is an eigenvalue of $A$ corresponding to the eigenvector which is in the span of $\{X, AX, A^{2} X, \dots , A^{k-1} X \}$ though $g(x)$ need not be an annihilating polynomial of $A$. But when $\deg\ g(x) = n$, I observed that $g(x)$ becomes characteristics polynomial of $A$ according to the example which I constructed.

But I don't understand why is this happening. Can we construct an example which disprove it or is it a true fact?

Please help me in understanding this fact.Thank you in advance.

Rodrigo de Azevedo
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Arnab Chattopadhyay.
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1 Answers1

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If $\deg g = n$, then $X$ is a cyclic vector for $A$.

It is well known that a linear transformation has a cyclic vector iff its minimal polynomial coincides with its characteristic polynomial.

lhf
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    See for instance http://math.stackexchange.com/questions/2025304/prove-that-t-has-a-cyclic-vector-iff-its-minimal-and-characteristic-polynomial. – lhf Dec 18 '16 at 18:50
  • That means @Ihf do you mean that there does not exist any polynomial $h(t)$ with $deg\ h(t) < n$ such that $h(t)$ annihilates $A$? – Arnab Chattopadhyay. Dec 18 '16 at 19:03
  • @ArnabChatterjee, yes. – lhf Dec 18 '16 at 20:14
  • Sorry @Ihf I have again a confusion.I have understood your claim and I have read the content in the link you provided.But how can I make sure that $g(t)$ is the minimal polynomial of $A$ whenever $\deg g = n$?Please help me with this concept. – Arnab Chattopadhyay. Dec 19 '16 at 03:57
  • So, we only have to show that $g(x)$ annihilates $A$ whenever $\deg g = n$. – Arnab Chattopadhyay. Dec 19 '16 at 05:36