A little more detail. The vectors are "dependent" if there exist number, , not all 0, such that . Since span the space, any vector, and in particular, the "w" vectors can be written as a linear combination of those: , , etc. Replacing the w vectors with those,
So
Now, we are NOT told that the u vectors are independent, only that they span the space. But certainly if the individual coefficients are 0 then certainly that sum will be the 0 vector. So one way that can happen is if
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We are given the "b" values, these are k equations for the l "a" values. If, as given, l>k, we have more unknown values than equations. We can take any values we want, in particular non- zero values, for l- k of them and have k equations to solve for the remaining k values.
Thats misleading. It implies you are elaborating on my proof, which is not the case. The detail applies to your proof, not mine. A better introduction to your post would be, A different proof which requires more detail.
By the way, there is nothing wrong with a different proof. It makes things interesting. Thanks
EDIT:
A simple example might make your proof easier to follow:
Let
k=1 and l=2.
Can we find a1 & a2 not both 0 st
a1w1+a2w2=0
Let
w1=b1u1, w2=b2u1 (given),
a1b1u1 + a2b2u1 = (a1b1+a2b2)u1=0 → a1b1+a2b2=0,
which can be solved for a1 & a2 not both 0, and w1 & w2 are therefore linearly dependent.
I missed a crucial point in the above proof, which I pass on in case anyone else did.
If
1) (a_{1}b_{11}+ +a_{l}b_{l1})u_{1}+ ..+ (a_{1}b_{1k}+ +a_{l}b_{lk})u_{k}=0,
which it is if the coefficients of u_{1}, ,u_{k}=0 , whether or not u_{1}, ,u_{k} are independent,
then
a_{1}(b_{11}u_{1}+ .+b_{1k}u_{k})+ +a_{l}(b_{l1}u_{1}+ +b_{lk}u_{k})=0, or
2) a_{1}w_{1}+a_{2}w_{2}+ ..a_{l}w_{l}=0
The point being, if you can solve 1) you can back-track to 2).
1) is the second eq after the first paragraph above.