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Exercise 3.D.7

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Proof. _(a)_ Let $S, T \in E$ . Then

0 = Sv + Tv = (S + T) v

Hence $E$ is closed under addition. Similarly, $E$ is closed under scalar multiplication. Therefore $E$ is indeed a subspace of $L (V, W)$ .

_(b)_ Let $U$ be a subspace of $V$ such that $V = U \oplus span (v)$ and let $u_{1}, \dots, u_{n}$ be a basis of $U$ . We will show that $E$ and $L (U, W)$ are isomorphic.

Define $Λ \in L (E, L (U, W))$ by

Λ (T) = T |_{U}

To prove $Λ$ is injective, let $S, T \in E$ such that $Λ (S) = Λ (T)$ . Let $u \in V$ . There are scalars $a_{1}, \dots, a_{n}, a \in 𝔽$ such that

u = a_{1} u_{1} + \dots + a_{n} u_{n} + av

Then

\begin{aligned} Tu & = T (a_{1} u_{1} + \dots + a_{n} u_{n} + av) \\ = a_{1} T u_{1} + \dots + a_{n} T u_{n} + aTv \\ = a_{1} T u_{1} + \dots + a_{n} T u_{n} \\ = a_{1} T |_{U} u_{1} + \dots + a_{n} T |_{U} u_{n} \\ = a_{1} S |_{U} u_{1} + \dots + a_{n} S |_{U} u_{n} \\ = a_{1} S u_{1} + \dots + a_{n} S u_{n} \\ = S (a_{1} u_{1} + \dots + a_{n} u_{n}) + S (av) \\ = S (u) \end{aligned}

Hence $S = T$ . Therefore $Λ$ is injective.

To prove surjectivity, let $T \in L (U, W)$ . Define $S \in L (V, W)$ by

\begin{aligned} S u_{j} & = T u_{j}, for j = 1, \dots, n \\ Sv & = 0 \end{aligned}

We have that $S \in E$ and it is easy to see that $Λ (S) = T$ . Hence $Λ$ is surjective, that is, an isomorphism between $E$ and $L (U, W)$ .

Therefore $\dim E = \dim U \dim W = (\dim V - 1) \dim W$ . □

celiopassos

2017-10-06 00:00

Exercise 3.D.7

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