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,
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
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.
By the way, there is nothing wrong with a different proof. It makes things interesting. Thanks
A simple example might make your proof easier to follow:
k=1 and l=2.
Can we find a1 & a2 not both 0 st
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.
1) (a1b11+ +albl1)u1+ ..+ (a1b1k+ +alblk)uk=0,
which it is if the coefficients of u1, ,uk=0 , whether or not u1, ,uk are independent,
a1(b11u1+ .+b1kuk)+ +al(bl1u1+ +blkuk)=0, or
2) a1w1+a2w2+ ..alwl=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.