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Exercise 9.A.2

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Proof. Let $u_{1}, u_{2}, v_{1}, v_{2} \in V$ . Then

\begin{aligned} T_{ℂ} ((u_{1} + i v_{1}) + (u_{2} + i v_{2})) & = T_{ℂ} ((u_{1} + u_{2}) + i (v_{1} + v_{2})) \\ = T (u_{1} + u_{2}) + iT (v_{1} + v_{2}) \\ = (T u_{1} + iT v_{1}) + (T u_{2} + iT v_{2}) \\ = T_{ℂ} (u_{1} + i v_{1}) + T_{ℂ} (u_{2} + i v_{2}) . \end{aligned}

Hence $T_{ℂ}$ satisfies the additivity property of linear maps. Now let $u, v \in V$ and $a, b \in ℝ$ . Then

\begin{aligned} T_{ℂ} ((a + bi) (u + iv)) & = T_{ℂ} ((au - bv) + i (av + bu)) \\ = T (au - bv) + iT (av + bu) \\ = (aTu - bTv) + i (aTv + bTu) \\ = (a + bi) (Tu + iTv) \\ = (a + bi) T_{ℂ} (u + iv) \end{aligned}

where the fourth line follows from the definition of complex scalar multiplication on $V_{ℂ}$ . Hence $T_{ℂ}$ satisfies the homogeneity property of linear maps. Therefore $T_{ℂ}$ is a linear map. □

celiopassos

2017-10-06 00:00

Exercise 9.A.2

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