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Let $A$ be an $n\times n$ matrix. Let $m >n$ such that $A^m=0$. Show that there exists an $r \leq n$ such that $A^r=0$.

Attempt at proof. Let $p(t)$ be a polynomial of degree $m$ such that $p(A)=0$. Let $p_A(t)$ be the characteristic polynomial of $A$. Then there exists $q_A(t)$ such that $q_A(A)=0$ and $r=\deg(q_A(t))\leq n$. Hence $A^r=0$.

Is this correct?

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What you write does not really make much sense. If you want to use the minimal polynomial and the Cayley-Hamilton theorem, you could argue as follows.

Since $A^m=0$, the minimal polynomial divides $X^m$, so it is of the form $X^r$ (by unique factorisation of polynomials, as $X$ is the only irreducible factor of$~X^m$). By the Cayley-Hamilton theorem, this minimal polynomial $X^r$ divides the characteristic polynomial, which is of degree$~n$. Therefore $r\leq n$.

But maybe you wanted to say the following. Let $p_A$ be the characteristic polynomial, then $p_A[A]=0$ by the Cayley-Hamilton theorem. Also $A^m=0$, so by dividing $p_A$ by $X^m$, one obtains a remainder $r_A$ such that $r_A[A]=0$, and $\deg r_A<n$. Now if $r=0$ it must be that $p_A=X^n$, and one can take $n=r$. However the the case $r_A\neq0$ it is not clear why $r_A$ should be of the form $X^r$; in fact it won't. (In fact one can show that $p_A=X^n$, so the other case does not occur, but I don't see how you could prove that more easily than what you need to prove.)

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The minimal polynomial divides all polynomials that vanish at $A$.

Therefore it divides $x^m$ (hence it is a power of $x$) and divides the characteristic polynomial. Since it divides the characteristic polynomial the minimal polynomial has degree smaller than or equal to the characteristic polynomial. The degree of the characteristic polynomial is $n$.

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