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Exercise 2.1.40

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1.

It’s linear since

η (u + v) = (u + v) + W = (u + W) + (v + W) = η (u) + η (v)

and

η (cv) = cv + W = c (v + W) = cη (v)

by the definition in Exercise 1.3.31. And for all element $v + W$ in $V ∕ W$ we have $η (v) = v + W$ and hence it’s onto. Finally if $η (v) = v + W = 0 + W$ we have $v - 0 = v \in W$ . Hence $N (η) = W$ .

2.

Since it’s onto we have

R (T) = V ∕ W

. And we also have

N (η) = W

. So by Dimension Theorem we have dim

(V) =

dim

(V ∕ W) +

dim

(W)

3.

They are almost the same but the proof in Exercise 1.6.35 is a special case of proof in Dimension Theorem.

jephianlin

2011-06-27 00:00

Exercise 2.1.40

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