# [Linear Algebra] Proof check. Nilpotent$\Rightarrow Spec\Rightarrow$ Characteristic Polynomial $\Rightarrow$ Nilpotent

Let $N$ be a linear operator over a finite-dimensional vector space $V$, over an algebraically closed field $K$. Show that the following are equivalent:

(1) $N$ is nilpotent.

(2) $Spec(N) = \{0\}$.

(3) The characteristic polynomial of $N$ is $c_N(x) = x^n$, where $n = dim(V)$.

My attempt:

We will show that $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$

$(1)\Rightarrow (2)$

$N$ is nilpotent, so $N^n=0$ for some $n$. Now let $v$ be an eigenvector: $Nv=?v$ for some scalar $?$. Now we get

$0=N^{n}v=?^{n}v ? ?=0$.

* I've shown by induction that $N^{n}v=?^{n}v$, just before. $K$ being algebraically closed was used here.

$(2)\Rightarrow (3)$

Knowing that all the eigenvalues of $N$ are 0 implies that the only root of the char. pol. is 0. Wich then tells us that the characteristic polynomial of $N$ is $c_N(x) = x^n$.

Is this too loose? I feel like I'm missing somehing here.

$(3)\Rightarrow (1)$

Knowing the char. pol., by the Cayley-Hamilton theorem, $N^n=0$.

Part of a worksheet for my Advanced Linear Algebra class. Is my proof correct? How would I be able to improve it? Any questions about notation or definitions just let me know.

* I've shown by induction that $N^{n}v=?^{n}v$, just before. $K$ being algebraically closed was used here.

$(2)\Rightarrow (3)$

Knowing that all the eigenvalues of $N$ are 0 implies that the only root of the char. pol. is 0. Wich then tells us that the characteristic polynomial of $N$ is $c_N(x) = x^n$.

Is this too loose? I feel like I'm missing somehing here.

$(3)\Rightarrow (1)$

Knowing the char. pol., by the Cayley-Hamilton theorem, $N^n=0$.

Part of a worksheet for my Advanced Linear Algebra class. Is my proof correct? How would I be able to improve it? Any questions about notation or definitions just let me know.

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