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Exercise:
Let mathcalUW subseteq V subspaces. Then: abcliste abc mathcalU+mathcalWtextSpmathcalUcapmathcalW. In particular mathcalU+mathcalWsubseteq V is a subspace and mathcalUmathcalWsubseteq mathcalU+mathcalW are subspaces of mathcalU+mathcalW. abc If mathcalUmathcalW are finite dimensional then so is mathcalU+mathcalW. Moreover one can find a basis for mathcalU+mathcalW as follows: itemize item choose a basis p_dotsp_k for mathcalUcapmathcalW where ktextdimmathcalUcapmathcalW. item Ext p_dotsp_k to a basis p_dotsp_ku_dotsu_l-k of mathcalU where ltextdimmathcalU. item Ext p_dotsp_k to a basis p_dotsp_kw_dotsw_m-k of mathcalW where mtextdimmathcalW. Then p_dotsp_ku_dotsu_l-kw_dotsw_m-k is a basis for mathcalU+mathcalW. itemize abc In particular we have textdimmathcalU+mathcalWtextdim mathcalU+textdimmathcalW-textdimmathcalUcapmathcalW. abcliste

Solution:
Proof. abcliste abc Clearly mathcalU+mathcalWsubseteq SpmathcalUcapmathcalW by the characterization of SpmathcalUcupmathcalW as the set of all linear combinations of elements from mathcalU cup mathcalW. But we also have SpmathcalU cup mathcalWsubseteq mathcalU+mathcalW. Indeed if v in SpmathcalUcupmathcalW then v _i^s a_iu_i+_j^r b_jw_j with a_ib_j in K u_iin U w_jin W. Longrightarrow vin mathcalU+mathcalW. This proves SpmathcalUcupmathcalWsubseteq mathcalU+mathcalW. Putting all together we have mathcalU+mathcalW SpmathcalUcapmathcalW. The rest of the statements in follow immediately. abc We first claim that p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Indeed if a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k+c_w_+...+c_m-kw_m-k. Write v:c_w_+...+c_m-kw_m-k-a_p_+...+_kp_k+b_u_+...+b_l-ku_l-k. Longrightarrow vinmathcalUcapmathcalW. Longrightarrow We can also write valpha_p_+...+alpha_kp_k for some alpha_...alpha_k in K. But vc_w_+...+c_m-kw_m-k. And p_...p_kw_...w_m-k are basis for mathcalW. Longrightarrow alpha_...alpha_k c_...c_m-k and v. But now v-a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k and p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Now clearly p_...p_ku_...u_l-kw_...w_m-k span mathcalU+mathcalW Longrightarrow p_...p_ku_...u_l-kw_...w_m-k is a basis for mathcalU+mathcalW. abc textdimmathcalU+mathcalWk+l-k+m-kl+m-ktextdimmathcalU+ textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
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Exercise:
Let mathcalUW subseteq V subspaces. Then: abcliste abc mathcalU+mathcalWtextSpmathcalUcapmathcalW. In particular mathcalU+mathcalWsubseteq V is a subspace and mathcalUmathcalWsubseteq mathcalU+mathcalW are subspaces of mathcalU+mathcalW. abc If mathcalUmathcalW are finite dimensional then so is mathcalU+mathcalW. Moreover one can find a basis for mathcalU+mathcalW as follows: itemize item choose a basis p_dotsp_k for mathcalUcapmathcalW where ktextdimmathcalUcapmathcalW. item Ext p_dotsp_k to a basis p_dotsp_ku_dotsu_l-k of mathcalU where ltextdimmathcalU. item Ext p_dotsp_k to a basis p_dotsp_kw_dotsw_m-k of mathcalW where mtextdimmathcalW. Then p_dotsp_ku_dotsu_l-kw_dotsw_m-k is a basis for mathcalU+mathcalW. itemize abc In particular we have textdimmathcalU+mathcalWtextdim mathcalU+textdimmathcalW-textdimmathcalUcapmathcalW. abcliste

Solution:
Proof. abcliste abc Clearly mathcalU+mathcalWsubseteq SpmathcalUcapmathcalW by the characterization of SpmathcalUcupmathcalW as the set of all linear combinations of elements from mathcalU cup mathcalW. But we also have SpmathcalU cup mathcalWsubseteq mathcalU+mathcalW. Indeed if v in SpmathcalUcupmathcalW then v _i^s a_iu_i+_j^r b_jw_j with a_ib_j in K u_iin U w_jin W. Longrightarrow vin mathcalU+mathcalW. This proves SpmathcalUcupmathcalWsubseteq mathcalU+mathcalW. Putting all together we have mathcalU+mathcalW SpmathcalUcapmathcalW. The rest of the statements in follow immediately. abc We first claim that p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Indeed if a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k+c_w_+...+c_m-kw_m-k. Write v:c_w_+...+c_m-kw_m-k-a_p_+...+_kp_k+b_u_+...+b_l-ku_l-k. Longrightarrow vinmathcalUcapmathcalW. Longrightarrow We can also write valpha_p_+...+alpha_kp_k for some alpha_...alpha_k in K. But vc_w_+...+c_m-kw_m-k. And p_...p_kw_...w_m-k are basis for mathcalW. Longrightarrow alpha_...alpha_k c_...c_m-k and v. But now v-a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k and p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Now clearly p_...p_ku_...u_l-kw_...w_m-k span mathcalU+mathcalW Longrightarrow p_...p_ku_...u_l-kw_...w_m-k is a basis for mathcalU+mathcalW. abc textdimmathcalU+mathcalWk+l-k+m-kl+m-ktextdimmathcalU+ textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
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eth, hs22, lineare algebra, proof
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(3, default)
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ENG (English)
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Proof
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