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Basis and linear dependence

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In physics exercises, we try to follow this pattern:

Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise

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Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise

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Level 5 -
Level 6 -

Exercise

https://texercises.com/exercise/basis-and-linear-dependence/

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Exercise:
Let v_...v_min V be a list of vectors which is linearly depent. Then exists leq j leq m: abcliste abc v_jin Spv_...v_j- abc Spv_...v_j-v_j+...v_mSpv_...v_m abcliste In other words we cna drop one of the vectors from the list without changing the span.

Solution:
Proof. Since the list is linearly depent exists a_...a_min K not all of them s.t. a_v_+...+a_mv_m Define j:textmaxi|a_ineq i.e. a_jneq a_j+...a_m. Longrightarrow a_v_+...+a_jv_j Longrightarrow v_jfrac-a_a_jv_+...+frac-a_j-a_jv_j- textok because a_jneq This proves a. To see b let vin Spv_...v_m. Longrightarrow v c_v_+...+c_mv_mquad textfor some c_...c_min K We can substitute o this and obtain that v can be written as a linear combination of v_...v_j-v_j+...v_m. This proves that textSpv_...v_m subseteq textSpv_...v_j-v_j+...v_m Obviously we also have textSpv_...v_j-v_j...v_msubseteq textSpv_..v_m Longrightarrow Spv_...v_j-v_j+...v_mSpv_...v_m

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Tags	basis, eth, hs22, lineare algebra, proof, span
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Difficulty	(3, default)
Points	0 (default)
Language	(English)
Type	Proof
Creator	rk
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