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Exercise 2.6.20

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1.

Assume that

T

is surjective. We may check whether

N (T^{t}) = {0}

or not. If

T^{t} (f) = fT = 0

, we have that

f (y) = f (T (x)) = 0

for all

y \in W

since there exist some

x \in V

such that

T (x) = y

. For the converse, assume that

T^{t}

is injective. Suppose, by contradiction,

R (T) \neq W

. By the previous exercise we can construct a nonzero linear functional

f (y) \in W^{∗}

such that

f (y) = 0

for all

y \in R (T)

. Let

f_{0}

be the zero functional in

W^{∗}

. But now we have that

T^{t} (f) (x) = f (T (x)) = 0 = T^{t} (g) (x)

, a contradiction. So

T

must be subjective.

2.

Assume that

T^{t}

is surjective. Suppose, by contradiction,

T (x) = 0

for some nonzero

x \in V

. We can construct a nonzero linear functional

g \in V^{∗}

such that

g (x) \neq 0

. Since

T^{t}

is surjective, we get some functional

f \in W^{∗}

such that

T^{t} (f) = g

. But this means

0 = f (T (x)) = T^{t} (f) (x) = g (x) \neq 0,

a contradiction.

For the converse, assume that $T$ is injective and let $S$ is a basis for $V$ . Since $T$ is injective, we have $T (S)$ is an independent set in $W$ . So we can extend it to be a basis $S^{'}$ for $W$ . Thus for every linear functional $g \in V^{∗}$ we can construct a functional $f \in W^{∗}$ such that $T^{t} (f) = g$ by the argument below. First we can construct a function $h \in F (S, 𝔽)$ by $h (T (s)) = g (s)$ for $s \in S$ and $h (t) = 0$ for all $t \in S^{'} ∖ T (S)$ . By Exercise 2.6.18 there is a linear functional $f \in W^{∗}$ such that $f_{S^{'}} = h$ . So now we have for all $s \in S$

g (s) = h (T (s)) = f (T (s)) = T^{t} (f) (s) .

By Exercise 2.1.34 we have $g = T^{t} (f)$ and get the desired conclusion.

jephianlin

2011-06-27 00:00

Exercise 2.6.20

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