Exercise
https://texercises.com/exercise/basis-linear-independence-generating/
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.

Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
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.
Meta Information
\(\LaTeX\)-Code
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.
Contained in these collections:

Attributes & Decorations
Tags
basis, eth, hs22, linear independence, lineare algebra, proof
Content image
Difficulty
(3, default)
Points
0 (default)
Language
ENG (English)
Type
Proof
Creator rk
Decoration
File
Link