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Exercise:
We define a map pi:Vlongrightarrow V/U by piv:vin V/Uquad forall vin V. Show that pi is a linear map. Moreover textImpiV/U i.e. pi is surjective and textKerpiU.

Solution:
Proof of linearity. pi v_+v_v_+v_v_+v_piv_+piv_ pi alpha v alpha valpha v alpha piv. Proof of surjectivity. Let xin V/U. By definition x is an equivalence class xv of some element vin V. So xpiv. Proof of Kerpisubseteq U. If vin textKerpi then piv &Longrightarrow v Longrightarrow vsim &Longrightarrow v-in U Longrightarrow vin U Proof of Usubseteq Kerpi. Let uin U &Longrightarrow usim quad textbecause &uu-in U Longrightarrow piuu
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Exercise:
We define a map pi:Vlongrightarrow V/U by piv:vin V/Uquad forall vin V. Show that pi is a linear map. Moreover textImpiV/U i.e. pi is surjective and textKerpiU.

Solution:
Proof of linearity. pi v_+v_v_+v_v_+v_piv_+piv_ pi alpha v alpha valpha v alpha piv. Proof of surjectivity. Let xin V/U. By definition x is an equivalence class xv of some element vin V. So xpiv. Proof of Kerpisubseteq U. If vin textKerpi then piv &Longrightarrow v Longrightarrow vsim &Longrightarrow v-in U Longrightarrow vin U Proof of Usubseteq Kerpi. Let uin U &Longrightarrow usim quad textbecause &uu-in U Longrightarrow piuu
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eth, hs22, lineare algebra, proof, quotient space
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(3, default)
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ENG (English)
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Proof
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