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frank0520
 one year ago
can there be two different linear combinations for the same vector?
frank0520
 one year ago
can there be two different linear combinations for the same vector?

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terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Sure. Just not of the basis.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2I mean... for instance, if we let \(\large V\) be a vector space with basis \(\large \{\vec{v_1},\vec{v_2},\vec{v_3}\}\) Now suppose we have a vector \[\large \vec u \in V\] Such that \(\large \vec u = a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}\)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0is this linear algebra, or before

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443248656889:dw

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Now suppose the vector \(\large \vec u\) had "another" linear combination, this time: \[\large \vec u = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Then since both of them are equal to \(\large \vec u\) It stands to reason that: \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}=\vec u = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Or... \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3} = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\] I just removed \(\large \vec u\) from the equation ^_^

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Now, using the addition property, we get: \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3} \left( b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3} \right)=\vec 0 \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0`look at the parallelogram rule for vector addition, that displays 2 different combinations to result in the sum` that won't work because the parallel sides are actually the same vector it's like adding in different orders a+b = b+a

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Let's "simplify" \[\large (a_1b_1)\vec{v_1}+(a_2b_2)\vec{v_2}+(a_3b_3)\vec{v_3}=\vec 0\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Now, do you remember that bases (plural of basis, anyone? haha) must be LINEAR INDEPENDENT? It means any linear combination of the basis that gives the zero vector must force all the coefficients to be zero ^_^ \[\Large a_1b_1 = 0\\\Large a_2b_2 = 0\\\Large a_3b_3 = 0\] Or in other words: \[\Large a_1=b_1\\\Large a_2=b_2 \\\Large a_3=b_3\] Means it must have been the same linear combination. Now if it wasn't the basis, or it wasn't a linear independent spanning set, things could be different ^^

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0does this mean the sketch is correct therefore solution is no?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0i will sit here and read while you type the entire course review,

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal no the sketch can be correct, different vectors, not parallelogram terenz showed that you can not use the Basis of the space

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0like <i, j , k> for 3 space

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0not a vector, well i guess it is maybe in 1D. lol
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