Exercise 1.6.20

Answers

1.: If $S = \emptyset$ or $S = {0}$ , then we have $V = {0}$ and the empty set can generate $V$ . Otherwise we can choose a nonzero vector $u_{1}$ in $S$ , and continuing pick $u_{k + 1}$ such that $u_{k + 1} \notin span ({u_{1}, u_{2}, \dots, u_{k}})$ . The process would teminate before $k > n$ otherwise we can find linearly independent set with size more than $n$ . If it terminates at $k = n$ , then we knoew the set is the desired basis. If it terminates at $k < n$ , then this means we cannot find any vector to be the vector $u_{k + 1}$ . So any vectors in $S$ is a linear combination of $β = {u_{1}, u_{2}, \dots, u_{k}}$ and hence $β$ can generate $V$ since $S$ can. But by Replacement Theorem we have $n \leq k$ . This is impossible.
2.: If $S$ has less than $n$ vectors, the process must terminate at $k < n$ . It’s impossible.

jephianlin

2011-06-27 00:00