Originally Posted by
HallsofIvy A little more detail. The vectors $\displaystyle \{w_1, w_2, ..., w_l\}$ are "dependent" if there exist number, $\displaystyle a_1, a_2,..., a_l$, not all 0, such that $\displaystyle a_1w_1+ a_2w_2+ ...+ a_lw_l= 0$. Since $\displaystyle \{u_1, u_2, ..., u_k\}$ span the space, any vector, and in particular, the "w" vectors can be written as a linear combination of those: $\displaystyle w_1= b_{11}u_1+ b_{12}u_2+ ...+ b_{1k}u_k$, $\displaystyle w_2= b_{21}u_1+ b_{22}u_2+ ...+ b_{2k}u_k}$, etc. Replacing the w vectors with those,
$\displaystyle a_1(b_{11}u_1+ b_{12}u_2+ ...+ b_{1k}u_k)+ a_2(b_{21}u_1+ b_{22}u_2+ ...+ b_{2k}u_k))+ ...+ a_l(b_{l1}u_1+ b_{l2}u_2+ ...+b_{lk}u_k)= 0$
So $\displaystyle (a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l})u_1+ (a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l})u_2+ ...+ (a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl})u_k= 0$
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
$\displaystyle a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l}= 0$
$\displaystyle a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l}= 0$
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$\displaystyle a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl}= 0$
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.