Orthogonal complement
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Exercise:
Let Phineq STsubseteq V. Then: abcliste abc S^perpsubseteq V is a linear subspace. abc ^perpV V^perp. abc Scap S^perp is either Phi or equal to . abc If Ssubseteq T then S^perpsupseteq T^perp. abc textSpS^perp S^perp abc Ssubseteq S^perp^perp abcliste
Solution:
abcliste abc in S^perp because langle srangle quad forall sin S. If alpha beta in K vwin S^perp then forall sin S we have langle alpha v+beta w srangle alphalangle vsrangle + betalangle wsrangle &Longrightarrow alpha v+beta vin S^perp. abc ^perpV. Let vin V^perpLongrightarrow langle vwrangle quad forall win V. In particular langle vvrangle Longrightarrow v. This shows V^perpsubseteq . But clearly subseteq V^perp. So V^perp. abc If Scap S^perp varnothing we are done. Otherwise ase that sin Scap S^perpLongrightarrow langle ssrangle Longrightarrow s. abc todo marked as exercise abc We have Ssubseteq textSpS. By d textSpS^perpsubseteq S^perp. We'll show that S^perpsubseteq textSpS^perp. Indeed let vin S^perp. Let uin textSpS. Write u_i^kalpha_iu_i with alpha_iin K u_iin S forall i. langle vurangle _i^k overlinealpha_ilangle vu_irangle Longrightarrow vperp u. This holds forall uin textSpS hence vin textSpS^perp. This shows S^perpsubseteq textSpS^perp. abc Let sin S. Then forall vin S^perp we have langle vsrangle Longrightarrow langle svrangle Longrightarrow sperp vquad forall vin S^perp Longrightarrow sin S^perp^perp. abcliste
Let Phineq STsubseteq V. Then: abcliste abc S^perpsubseteq V is a linear subspace. abc ^perpV V^perp. abc Scap S^perp is either Phi or equal to . abc If Ssubseteq T then S^perpsupseteq T^perp. abc textSpS^perp S^perp abc Ssubseteq S^perp^perp abcliste
Solution:
abcliste abc in S^perp because langle srangle quad forall sin S. If alpha beta in K vwin S^perp then forall sin S we have langle alpha v+beta w srangle alphalangle vsrangle + betalangle wsrangle &Longrightarrow alpha v+beta vin S^perp. abc ^perpV. Let vin V^perpLongrightarrow langle vwrangle quad forall win V. In particular langle vvrangle Longrightarrow v. This shows V^perpsubseteq . But clearly subseteq V^perp. So V^perp. abc If Scap S^perp varnothing we are done. Otherwise ase that sin Scap S^perpLongrightarrow langle ssrangle Longrightarrow s. abc todo marked as exercise abc We have Ssubseteq textSpS. By d textSpS^perpsubseteq S^perp. We'll show that S^perpsubseteq textSpS^perp. Indeed let vin S^perp. Let uin textSpS. Write u_i^kalpha_iu_i with alpha_iin K u_iin S forall i. langle vurangle _i^k overlinealpha_ilangle vu_irangle Longrightarrow vperp u. This holds forall uin textSpS hence vin textSpS^perp. This shows S^perpsubseteq textSpS^perp. abc Let sin S. Then forall vin S^perp we have langle vsrangle Longrightarrow langle svrangle Longrightarrow sperp vquad forall vin S^perp Longrightarrow sin S^perp^perp. abcliste