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anonymous
 5 years ago
Linear Algebra: Consider orthogonal unit vectors v1>vm in Rn. Show they are necessarily linearly independent. Says to use dot product in proof, but not really sure of an attack strategy here
anonymous
 5 years ago
Linear Algebra: Consider orthogonal unit vectors v1>vm in Rn. Show they are necessarily linearly independent. Says to use dot product in proof, but not really sure of an attack strategy here

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The basic way to show vectors are independent is to show their linear combination is 0, and the only such coefficients are all 0. So \[c_1v_1 + c_2v_2 + ... + c_mv_m = 0 \] and we aim to prove all c_n are 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From the above, we can take any one of {v1, v2...vm}, and dot it with both sides: \[v_i \cdot (c_1v_1 + c_2v_2 + ... + c_mv_m) = v_i \cdot 0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The dot product is distributive, so we can dot v_i with each term separately, and on the right side we have 0. Once we distribute v_i throughout, however, every term becomes 0, except the term that matches the vector we used to dot it with: \[v_i \cdot (c_1v_1 + c_2v_2 + ... + c_mv_m) = 0\] \[c_iv_i \cdot v_i\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(forgot the "= 0" above) The above happens because all vectors are orthogonal, and two orthogonal vectors' dot product are 0, and v_i dot v_i is nonzero, so c_i must be 0. This can be repeated with all vectors, yielding all c are 0, and that's the proof.
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