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Exercise 3.B.23

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Proof. Because $range ST \subset range S$ , it follows that $\dim range ST \leq \dim range S$ . Let $v_{1}, \dots, v_{n}$ be a basis of $range T$ . It is easy to see that $range ST = span (S v_{1}, \dots, S v_{n})$ . Hence

\dim range ST = \dim span (S v_{1}, \dots, S v_{n}) \leq n = \dim range T

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celiopassos

2017-10-06 00:00

Exercise 3.B.23

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