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Exercise:
Let V be a finite dimensional vector space and mathcalUsubseteq V a subspace. Then exists a subspace mathcalWsubseteq V which is a complement to mathcalU.

Solution:
Proof. Choose a basis u_...u_l to mathcalU where l dim l. Ext this basis to a basis of V: u_...u_lw_...w_m where l+mdimV. Take mathcalW:Spw_...w_m. Clearly mathcalU+mathcalWV. And mathcalUcapmathcalW because if _i^l a_iu_i_j^m b_jw_j Longrightarrow _i^l a_iu_i+_j^m -b_jw_j But u_...u_lw_...w_m are linearly indepent Longrightarrow a_...a_lb_...b_m Longrightarrow v.
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Exercise:
Let V be a finite dimensional vector space and mathcalUsubseteq V a subspace. Then exists a subspace mathcalWsubseteq V which is a complement to mathcalU.

Solution:
Proof. Choose a basis u_...u_l to mathcalU where l dim l. Ext this basis to a basis of V: u_...u_lw_...w_m where l+mdimV. Take mathcalW:Spw_...w_m. Clearly mathcalU+mathcalWV. And mathcalUcapmathcalW because if _i^l a_iu_i_j^m b_jw_j Longrightarrow _i^l a_iu_i+_j^m -b_jw_j But u_...u_lw_...w_m are linearly indepent Longrightarrow a_...a_lb_...b_m Longrightarrow v.
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eth, hs22, linear subspace, lineare algebra, proof, vector space
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(3, default)
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ENG (English)
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Proof
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