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Exercise 3.D.2

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Proof. Let $N$ be the set of noninvertible operators on $V$ . Let $v_{1}, \dots, v_{n}$ be a basis of $V$ . Define $S, T \in L (V)$ such that

\begin{aligned} S (a_{1} v_{1} + \dots + a_{n} v_{n}) & = a_{1} v_{1} \\ T (a_{1} v_{1} + \dots + a_{n} v_{n}) & = a_{2} v_{2} + \dots + a_{n} v_{n} \end{aligned}

Neither $S$ and $T$ are surjective, hence both are in $N$ . But $S + T = I$ and $I$ is clearly invertible. Thus $S + T \notin N$ and $N$ is not closed under addition. □

celiopassos

2017-10-06 00:00

Exercise 3.D.2

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