Prove $0$ is the only eigenvalue of $A$.

Question

Let $A$ be an $n$ x $n$ real matrix. Prove $0$ is eigen value of $A$.

PROOF:

Let $m∈ℤ^{+}$

so $A^{m}$x$=λ^{m}$x

there exists$m≥1$ s.t $A^{m}=0$.

$0=λ^{m}$x, but $x≠0$, so $λ^{m}=0$

$m>0$ , so $λ^{m}=0$ IFF so $λ=0$.

so all $λ=0$

Answer 1

Here is a rewrite of your proof that I think is an improvement.

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Proof: We are given that $A$ is $n \times n$, and that $A$ is nilpotent. Let $m$ denote the smallest positive integer for which $A^m = 0$.

First, we show that $A$ must have $0$ as an eigenvalue. Let $\mathbf x \in \Bbb R^n$ be any non-zero vector. Let $j_0$ denote the smallest positive integer $j$ for which $A^j\mathbf x = 0$; we note that such an integer exists because taking $j = m$ yields $A^j\mathbf x = 0 \mathbf x = 0$. We can see that $A(A^{j_0 - 1} \mathbf x) = 0 \cdot \mathbf x$, and $A^{j_0 - 1} \mathbf x \neq \mathbf 0$. So, $A^{j_0 - 1}\mathbf x$ is an eigenvector of $A$ associated with $\lambda = 0$.

Now, suppose that $\lambda$ is an eigenvector of $A$. Let $\mathbf x \in \Bbb R^n$ be an associated eigenvector so that $\mathbf x \neq \mathbf 0$ and $A \mathbf x = \lambda \mathbf x$.

Claim: $A^k \mathbf x = \lambda^k \mathbf x$.

Proof of claim: We proceed by induction

Base case: $A^1 \mathbf x= \lambda^1 \mathbf x$.

Inductive step: If $A^n \mathbf x = \lambda^n \mathbf x$, then $$ A^{n+1} \mathbf x = A (A^n \mathbf x) = A \lambda^n \mathbf x = \lambda^n A \mathbf x = \lambda^n \cdot \lambda \mathbf x = \lambda^{n+1} \mathbf x. \quad \square $$

We know that $A^m = 0$. By the claim, it follows that $$ \mathbf 0 = 0 \cdot \mathbf x = A^m \mathbf x = \lambda^m \mathbf x. $$ If $\lambda^m \mathbf x = 0$, then we must have $\lambda^m = 0$. However, $m$ is a positive integer, so $\lambda^m = 0$ implies that $\lambda = 0$.

That is, if $\lambda$ is an eigenvalue of $A$, then $\lambda = 0$. So, $\lambda = 0$ is the only eigenvalue of $A$.