Basis, linear independence, generating
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Exercise:
Let V be a vector space of dimVnin mathbbZ_geq . Let v_...v_n in V. Then the following statements are equivalent: abcliste abc v_...v_n are linearly indepent. abc v_...v_n generate V i.e. Spv_...v_nV. abc v_...v_n is a basis for V. abcliste
Solution:
Proof Longrightarrow . By the mary above we can ext the list v_...v_n to a basis. But v_...v_n has already length ndimV Longrightarrow v_...v_n is already a basis. Proof Longrightarrow . This is part of the definition of a basis. Proof Longrightarrow . If v_...v_n generate V then by the mary above a sublist of v_...v_n is a basis. But v_...v_n contains ndimV elements Longrightarrow This is already a basis.
Let V be a vector space of dimVnin mathbbZ_geq . Let v_...v_n in V. Then the following statements are equivalent: abcliste abc v_...v_n are linearly indepent. abc v_...v_n generate V i.e. Spv_...v_nV. abc v_...v_n is a basis for V. abcliste
Solution:
Proof Longrightarrow . By the mary above we can ext the list v_...v_n to a basis. But v_...v_n has already length ndimV Longrightarrow v_...v_n is already a basis. Proof Longrightarrow . This is part of the definition of a basis. Proof Longrightarrow . If v_...v_n generate V then by the mary above a sublist of v_...v_n is a basis. But v_...v_n contains ndimV elements Longrightarrow This is already a basis.