Exercise 4.1.6

Question

An operator $A$ is called 	extit{nilpotent} if $A^k = \mathbf{0}$ for some $K$. Prove that if $A$ is nilpotent, then $\sigma(A) = \{0\}$ (i.e. that 0 is the only eigenvalue of $A$).

Jing Huang · Accepted Answer

\begin{proof}
Note that if $\lambda$ is a nonzero eigenvalue of $A$ and $A \mathbf{x} = \lambda \mathbf{x}$. Then $A^2 \mathbf{x} = A(\lambda \mathbf{x}) = \lambda ^2 \mathbf{x}$, $A^3 \mathbf{x} = A(\lambda ^2 \mathbf{x}) = \lambda ^3 \mathbf{x}$ ... $A^k \mathbf{x} = \lambda ^k \mathbf{x}$. That is to say if $\lambda \in \sigma (A), \lambda^k \in \sigma (A^k)$. Now $A^k = \mathbf{0}$, $\sigma(A^k) = \{ 0 \}$. Then 0 is the only eigenvalue of $A$.
\end{proof}

Exercise 4.1.6

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Exercise 4.1.6

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