Exercise 8.A.8

Answers

Proof. False. Let $V = ℂ^{2}$ . Define $S, T \in L (ℂ)$ by

\begin{aligned} S (z_{1}, z_{2}) & = (0, z_{1}) \\ T (z_{1}, z_{2}) & = (z_{2}, 0) . \end{aligned}

Both $S$ and $T$ are nilpotent, however, $S + T$ is not (its square equals the identity). □

celiopassos

2017-10-06 00:00