Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 3.F.30

Exercise 3.F.30

Answers

Proof. We have

\begin{aligned} \dim (null φ_{1} \cap \dots \cap null φ_{m}) & = \dim V - \dim ({(null φ_{1} \cap \dots \cap null φ_{m})}^{0}) \\ = \dim V - \dim ({(null φ_{1})}^{0} + \dots + {(null φ_{m})}^{0}) \\ = \dim V - \dim (span (φ_{1}) + \dots + span (φ_{m})) \\ = \dim V - \dim (span (φ_{1}, \dots, φ_{m})) \\ = \dim V - m \end{aligned}

Where the first equation follows from 3.106, the second from Exercise 23, the third from Theorem 1 in Chapter 3 notes, the fourth from the definition of the sum of subspaces and the fifth because $φ_{1}, \dots, φ_{m}$ is linearly independent. □

celiopassos

2017-10-06 00:00

Exercise 3.F.30

Answers

Comments

Add answer