Exercise 1.6.35

Answers

1.

Since

{u_{1}, u_{2}, \dots, u_{n}}

is a basis, the linear combination of

{u_{k + 1}, u_{k + 2}, \dots, u_{n}}

can not be in span $({u_{1}, u_{2}, \dots, u_{k}}) = W$ . This can make sure that

{u_{k + 1} + W, u_{k + 2} + W, \dots, u_{n} + W}

is linearly independent by 1.3.31(b). For all $u + W \in V ∕ W$ we can write

\begin{array}{l} u + W & = (a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}) + W \\ = (a_{1} u_{1} + a_{2} u_{2} + \dots + a_{k} u_{k}) + (a_{k + 1} u_{k + 1} + a_{k + 2} u_{k + 2} + \dots + a_{n} u_{n}) + W \\ = (a_{k + 1} u_{k + 1} + a_{k + 2} u_{k + 2} + \dots + a_{n} u_{n}) + W \\ = a_{k + 1} (u_{k + 1} + W) + a_{k + 2} (u_{k + 2} + W) + \dots + a_{n} (u_{n} + W) \end{array}

and hence it’s a basis.

2.

By preious argument we have dim

(V ∕ W) = n - k =

dim

(V) -

dim

(W)

jephianlin

2011-06-27 00:00