Exercise 2.4.17

Answers

1.

y_{1}, y_{2} \in T (V_{0})

and

y_{1} = T (x_{1})

y_{2} = T (x_{2})

, we have that

y_{1} + y_{2} = T (x_{1} + x_{2}) \in T (V_{0})

and

c y_{1} = T (c x_{1}) = T (V_{0})

. Finally since

V_{0}

is a subspace and so

0 = T (0) \in T (V_{0})

T (V_{0})

is a subspace of

W

2.

We can consider a mapping

T^{'}

from

V_{0}

T (V_{0})

T^{'} (x) = T (x)

for all

x \in V_{0}

. It’s natural that

T^{'}

is surjective. And it’s also injective since

T

is injective. So by Dimension Theorem we have that

\dim (V_{0}) = \dim (N (T^{'})) + \dim (R (T^{'})) = \dim (T (V_{0})) .

jephianlin

2011-06-27 00:00