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

1. DanJS

mhmmm

2. terenzreignz

Sure. Just not of the basis.

3. terenzreignz

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

is this linear algebra, or before

5. triciaal

|dw:1443248656889:dw|

6. terenzreignz

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

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

@jim_thompson5910

9. terenzreignz

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

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

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

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

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

does this mean the sketch is correct therefore solution is no?

15. DanJS

good one @terenzreignz

16. DanJS

i will sit here and read while you type the entire course review,

17. DanJS

@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

like <i, j , k> for 3 space

19. zzr0ck3r

1+3=4 2+2=4 :)

20. DanJS

haha, yep

21. DanJS

not a vector, well i guess it is maybe in 1D. lol

22. DanJS

the direction is + or -