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Consider any matrix $A$ of order $n\times n.$ How to show that any monic polynomial of degree $n$ whose root is $A$ is the characteristic polynomial of $A?$

I have used the result several times without knowing the proof.

Thomas Andrews
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Sriti Mallick
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    This is not true. For instance, if $A=0$ is $5\times 5$. Then $X^5$ and $X^5-42X^3$ both annihilate $A$. – Julien May 15 '13 at 13:48
  • @julien: What about the nonzero matrices? – Sriti Mallick May 15 '13 at 13:51
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    The same applies to a nilpotent matrix such that $A^3=0$. But more interestingly, if the degree of the minimal polynomial $m(x)$ is less than the degree of the characteristic polynomial $p(x)$, any $p(x)-\lambda m(x)$ will do. – Julien May 15 '13 at 14:08

2 Answers2

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This result is true if and only if the minimal and characteristic polynomial of $A$ are the same. It is true for example if $A$ has distinct eigenvalues...

N. S.
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This works when the minimal polynomial of $A$, $m_A(x)$, is $n$th order.

Why? Because given another $n$th degree polynomial $f$ with leading coefficient 1, $f-m_A$ has $A$ as a root but has degree lower than $n$. Therefore it must be 0.

not all wrong
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