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

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

Mathematics
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mhmmm
Sure. Just not of the basis.
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 \(\large \vec u = a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}\)

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is this linear algebra, or before
|dw:1443248656889:dw|
Now suppose the vector \(\large \vec u\) had "another" linear combination, this time: \[\large \vec u = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\]
Then since both of them are equal to \(\large \vec u\) It stands to reason that: \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}=\vec u = b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3}\]
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 ^_^
Now, using the addition property, we get: \[\large a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3} -\left( b_1\vec{v_1}+b_2\vec{v_2}+b_3\vec{v_3} \right)=\vec 0 \]
`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
Let's "simplify" \[\large (a_1-b_1)\vec{v_1}+(a_2-b_2)\vec{v_2}+(a_3-b_3)\vec{v_3}=\vec 0\]
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 ^^
does this mean the sketch is correct therefore solution is no?
good one @terenzreignz
i will sit here and read while you type the entire course review,
@triciaal no the sketch can be correct, different vectors, not parallelogram terenz showed that you can not use the Basis of the space
like for 3 space
1+3=4 2+2=4 :)
haha, yep
not a vector, well i guess it is maybe in 1D. lol
the direction is + or -

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