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

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$[Linear Algebra] Proof check. Nilpotent$\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: $Nv=λv$ for some scalar $λ$. Now we get

$0=N^{n}v=λ^{n}v ⇒ λ=0$.

* I've shown by induction that $N^{n}v=λ^{n}v$, 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.

Linear Algebra Vector Spaces Eigenvalues & Eigenvectors

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I'd say your proof is totally fine, I don't think it needs any improvement. If you really want to be extra precise, you can use only the definition of characteristic polynomial as $\det(xI-N)$ (or maybe $N-xI$, depending on how you defined it), and rewrite $(2) \implies (3)$ as follows.

Since the only eigenvalue is $0$ and the characteristic polynomial is invariant under conjugation, you can assume that $N$ is in normal Jordan form, or, if you don't have Jordan form, that it can be made upper triangular with the eigenvalues (i.e. all $0$) on the diagonal. Then $xI-N$ is upper triangular with all $x$ on the diagonal, and so its determinant is just the product of the elements on the diagonal, which is obviously $x^n$, $n$ being the size of the matrix (or the dimension of $V$).

But I think your proof is also perfectly fine, just mention that the roots of the characteristic polynomial are exactly the eigenvalues of $N$. If I was grading it I would give it a perfect score.

Alessandro Iraci

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

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