[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.
* 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.
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.
![Alessandro Iraci](https://matchmaticians.com/storage/user/100977/thumb/matchmaticians-ee0kis-file-1-avatar-512.jpg)
1.7K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 620 views
- $4.00
Related Questions
- Hello! I Would like a proof detailed of the following question.
- Find the null space of the matrix $\begin{pmatrix} 1 & 2 & -1 \\ 3 & -3 & 1 \end{pmatrix}$
- Find $x$ so that $\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible
- Advice for proving existence claims
- Consider the function, prove that it's bilinear, symmetric, and positive definite
- The Span and Uniqueness of Solutions in a Parametric Matrix
- Find $x$ so that $\begin{bmatrix} 2 & 0 & 10 \\ 0 & x+7 & -3 \\ 0 & 4 & x \end{bmatrix} $ is invertible
- Find eigenvalues and eigenvectors of $\begin{pmatrix} -3 & 0 & 2 \\ 1 &-1 &0\\ -2 & -1& 0 \end{pmatrix} $