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Exercise:
Let v_...v_min V be a list of vectors which is linearly depent and let v_...v_k be linearly indepent where leq jleq m is some index. Then k j.

Solution:
Proof. Ase by contradiction that jleq k. We have v_ja_v_+...+a_j-v_j- for some a_...a_j-in K. Longrightarrow a_v_+...+a_j-v_j-+-v_j But this is a non-trivial linear combination of v_...v_j that gives . Longrightarrow v_...v_jquad textare linearly depent Longrightarrow v_..v_k are also linearly depent because jleq k
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Exercise:
Let v_...v_min V be a list of vectors which is linearly depent and let v_...v_k be linearly indepent where leq jleq m is some index. Then k j.

Solution:
Proof. Ase by contradiction that jleq k. We have v_ja_v_+...+a_j-v_j- for some a_...a_j-in K. Longrightarrow a_v_+...+a_j-v_j-+-v_j But this is a non-trivial linear combination of v_...v_j that gives . Longrightarrow v_...v_jquad textare linearly depent Longrightarrow v_..v_k are also linearly depent because jleq k
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basis, eth, hs22, linear independence, lineare algebra, proof, span
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(3, default)
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Language
ENG (English)
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Proof
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