1. ## Linear algebra

Can someone prove this claim?

Claim:if w1,w2,.....,wl​ belongs to the span(u1,u2,....,uk) and l>k then w1,w2,.....,wl​ is linearly dependent.

2. ## Re: Linear algebra

Originally Posted by davidciprut
Claim:if w1,w2,.....,wl​ belongs to the span(u1,u2,....,uk) and l>k then w1,w2,.....,wl​ is linearly dependent.
dim span of u1...uk <= k
if w1...wl linearly independent,
dim span of u1...uk > k

3. ## Re: Linear algebra

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
$\displaystyle a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l}= 0$
$\displaystyle a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l}= 0$
.
.
.
$\displaystyle a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl}= 0$

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.

4. ## Re: Linear algebra

Originally Posted by HallsofIvy
A little more detail. ....
Thats 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.

5. ## Re: Linear algebra

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
$\displaystyle a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l}= 0$
$\displaystyle a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l}= 0$
.
.
.
$\displaystyle a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl}= 0$

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.
I missed a crucial point in the above proof, which I pass on in case anyone else did.

If
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,
then
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.