Prove that the characteristic polynomial of a nilpotent matrix is $x^n$ 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 $det(A)=0$.
 A: 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 $\lambda^n$.
A: 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 $x^n$
A: 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 $n$.
A: *

*Step. Two matrices A and B have the same characteristic polynomial, if there exists an invertible matrix S, such that $SAS^{-1}=B$:
$det(B-\lambda E_n)=det(SAS^{-1}-S\lambda S^{-1})=det(S(B-\lambda E_n)S^{-1})=det(B-\lambda E_n)$

*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 $\lambda^n$
