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How can I prove that the char.pol. of a nilpotent matrix is of the form $x^k$? I'm trying to do it by contradiction but assuming that $p_{xA}=a_0+a_1x+\dots+a_mx^m+\dots+a_nx^n$ seems not giving any contradiction. I've proved that the eigenvalues of A has to be 0, which led to $det(A)=0$.

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    $\begingroup$ If you've proven that all the eigenvalues of $A$ are 0, what does that tell you about the char. pol.? $\endgroup$ – Mathmo123 Jun 25 '14 at 20:57
  • $\begingroup$ That its only root is 0. $\endgroup$ – Allonsy Jun 25 '14 at 20:58
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    $\begingroup$ And what polynomial can you think of whose only root is zero? $\endgroup$ – Mathmo123 Jun 25 '14 at 21:00
  • $\begingroup$ Yes, $x^n$, but isn't there any more "formal" proof? $\endgroup$ – Allonsy Jun 25 '14 at 21:01
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    $\begingroup$ Ah I see. That's a good question. You may have noticed in your proof that the eigenvalues have to be 0 that it does not matter whether A is complex or real. If $A^k = 0$ in $\mathbb R$, then this will also be true in $\mathbb C$. If the char pol were $x(x^2 + 1)$, then we would have a non-zero eigenvalue in $\mathbb C$ $\endgroup$ – Mathmo123 Jun 25 '14 at 21:10
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As discussed in the comments, you can use the fact that the only eigenvalue is $0$. This means that the only root of the characteristic polynomial is $0$.

However, I do think that the exchange in the comments "Can you think of a polynomial whose only root is $0$?" "Yes, $x^n$." is insufficient. Just because polynomials of the form $x^n$ have $0$ as their only root does not mean that they're the only such polynomials.

You just need to say that any complex polynomial splits as $\prod(x-r_i)$ where the $r_i$ are the roots of the polynomial, and since in this case the $r_i$ are all $0$, we have $\prod x=x^n$ for some $n$.

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  • $\begingroup$ Ultimately that is a question of how much you are allowed to assume. I could equally challenge your answer and say that you need to prove that any complex polynomial splits in that way. And I could keep going. Hey while we're at it, let's start from the basic axioms. It's a generally accepted rule that we can quote basic results from other branches of mathematics without proof. Given that we are working in linear algebra and not basic function theory, I think its perfectly acceptable to assume this fact without proof. $\endgroup$ – Mathmo123 Jun 25 '14 at 21:27
  • $\begingroup$ @Mathmo123 I don't think it's acceptable at all. Maybe $x^5+6x+75x-2x$ has no roots other than the obvious root of $0$. I don't know, it's very hard to tell at a glance. You need the fact that you can split polynomials. You have to at least mention that very non-obvious fact, otherwise what you have doesn't even come close to being a proof. $\endgroup$ – Jack M Jun 26 '14 at 9:19
  • $\begingroup$ I think that your complaint is that I'm being too rigid, and that we should be able to accept intuitively obvious facts as intuitively obvious without proof. What I'm saying is that if you don't know about factorization, the fact that $x^n$ is the only polynomial whose only root of $0$ isn't intuitively obvious. $\endgroup$ – Jack M Jun 26 '14 at 9:26
  • $\begingroup$ I see your argument. However it is also far from obvious that any complex polynomial splits - by your own argument, we should have to prove that too... and so on until the axioms. Indeed, given that we could be working in the real and not the complex case, that argument is actually very dubious - in the real case it is NOT true that every polynomial splits. The point I'm making is that since this is a proof from another area of mathematics, it's fair enough to quote this result. $\endgroup$ – Mathmo123 Jun 26 '14 at 18:18
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If $A^k = 0$, then $(\lambda I - A)(\lambda^{k - 1}I + \lambda^{k - 2}A + \cdots + A^{k - 1}) = \lambda^kI$. Taking determinants, we get that the characteristic polynomial of $A$ divides $det(\lambda^kI) = \lambda^{nk}$. Since the characteristic polynomial of $A$ is a monic polynomial of degree $n$, it follows that it must be $\lambda^n$.

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Suppose $A^k = 0$, and $A^{s} \neq 0$ if $s < k$.

Then suppose that $\lambda \neq 0$ is an eigenvalue. Then there exists $x \neq 0$ so that $Ax = \lambda x$. Multiply both sides by $A^{k-1}$ to get $A^k x = \lambda A^{k-1}x$. But the left hand side is $0$, while the right hand side is not.

This means that the only eigenvalue can be $\lambda = 0$, and this implies that the charachteristic polynomial is $x^n$

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  • $\begingroup$ How do you guarantee that $A^{k-1}x\neq0$? $\endgroup$ – Majid May 22 '19 at 13:57
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  1. Step. Two matrices A and B have the same characteristic polynomial, if there exists an invertible matrix S, such that $SAS^{-1}=B$:

    $det(B-\lambda E_n)=det(SAS^{-1}-S\lambda S^{-1})=det(S(B-\lambda E_n)S^{-1})=det(B-\lambda E_n)$

  2. Step: Show that for every nilpotent matrix A there exists an invertible matrix S, such that: $A=SDS^{-1}$

$$ D=\begin{matrix} 0 & * & \ldots & *\\ 0 & 0 & \ldots & *\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\ldots & 0 \end{matrix} $$

Instructions for the prove of 2:

n ist the first natural number such that $A^n$=0

1) For $V_k:=ImageA^{d-k}$ show the following:

${0}=V_0 \subseteq V_1 \subseteq V_2 \subseteq...\subseteq V_{d-1} \subseteq V_d =\mathbb R^n$

2) Choose a basis of $V_1$ and then expand it to a basis of $V_2$ and so on...until u get a basis of $\mathbb R^n$

3) Show then that A looks like above in that basis.

And now u can conclude that the characteristic polynomial looks like $\lambda^n$

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