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frank0520

  • one year ago

can there be two different linear combinations for the same vector?

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  1. DanJS
    • one year ago
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    mhmmm

  2. terenzreignz
    • one year ago
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    Sure. Just not of the basis.

  3. terenzreignz
    • one year ago
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    I 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 \(\large \vec u = a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}\)

  4. DanJS
    • one year ago
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    is this linear algebra, or before

  5. triciaal
    • one year ago
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    |dw:1443248656889:dw|

  6. terenzreignz
    • one year ago
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    Now suppose the vector \(\large \vec u\) had "another" linear combination, this time: \[\large \vec u = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\]

  7. terenzreignz
    • one year ago
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    Then since both of them are equal to \(\large \vec u\) It stands to reason that: \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}=\vec u = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\]

  8. triciaal
    • one year ago
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    @jim_thompson5910

  9. terenzreignz
    • one year ago
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    Or... \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3} = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\] I just removed \(\large \vec u\) from the equation ^_^

  10. terenzreignz
    • one year ago
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    Now, using the addition property, we get: \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3} -\left( b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3} \right)=\vec 0 \]

  11. jim_thompson5910
    • one year ago
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    `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

  12. terenzreignz
    • one year ago
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    Let's "simplify" \[\large (a_1-b_1)\vec{v_1}+(a_2-b_2)\vec{v_2}+(a_3-b_3)\vec{v_3}=\vec 0\]

  13. terenzreignz
    • one year ago
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    Now, 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 ^_^ \[\Large a_1-b_1 = 0\\\Large a_2-b_2 = 0\\\Large a_3-b_3 = 0\] 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 ^^

  14. triciaal
    • one year ago
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    does this mean the sketch is correct therefore solution is no?

  15. DanJS
    • one year ago
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    good one @terenzreignz

  16. DanJS
    • one year ago
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    i will sit here and read while you type the entire course review,

  17. DanJS
    • one year ago
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    @triciaal no the sketch can be correct, different vectors, not parallelogram terenz showed that you can not use the Basis of the space

  18. DanJS
    • one year ago
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    like <i, j , k> for 3 space

  19. zzr0ck3r
    • one year ago
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    1+3=4 2+2=4 :)

  20. DanJS
    • one year ago
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    haha, yep

  21. DanJS
    • one year ago
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    not a vector, well i guess it is maybe in 1D. lol

  22. DanJS
    • one year ago
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    the direction is + or -

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