# One set v is a linear combination of u. Prove u is linearly dependent

• cbarker1

#### cbarker1

Gold Member
MHB
Homework Statement
Let ##\{u_1,\ldots,u_s\}##and ##\{v_1,\ldots,v_t\}## be two sets of vectors. If ##s>t## and ##u_i## is a linear combination of ##v_1,\ldots, v_t##, show ##u_1,\ldots,u_s## is linearly dependent
Relevant Equations
Linear combination: c1u1+c2u2+c3u3+...+cnun=0
Linearly dependent is when there is a nontrival solution of the linear combination.
Hi Everybody,

I am having some difficulties on the prove this problem.
I picked a nice example when I was trying to think about the proof.

Let ##s=3## and ##t=2##. Then ##u1=c1v1+c2v2, u2=c3v1+c4v2, u3=c5v1+c6v2##. Then a linear combination of u: ##K1u1+K2u2+K3u3=0##. I grouped both linear combination of u in terms of v:
##K1(c1v1+c2v2)+K2(c3v1+c4v2)+K3(c5v1+c6v2)=0## It implies this system of linear equations for c1,..c6:
(K1c1+c3K2+c5K3)=0
(K1c2+c4K2+K3c6)=0
Here is where I am lost. What should I do next? Solve for K1,K2,K3? or something else.

Will this example help me prove this exercise?

Thanks,
cbarker 1

Let ##s=3## and ##t=2##
This is strange. In the problem statement it says 'If ##s<t## ' ?

• FactChecker
What if ##s=1 < 2=t## and ##u_1## is a straight in the plane spanned by ##\{v_1,v_2\}##. Then ##K\cdot u_1=0## implies ##K=0## and ##\{u_1\}## is linear independent.

It seems quite obvious that the problem statement is wrongly reproduced and should say ##s>t## instead of ##s<t##.

Homework Statement:: Let ##\{u_1,\ldots,u_s\}##and ##\{v_1,\ldots,v_t\}## be two sets of vectors. If ##s<t## and ##u_i## is a linear combination of ##v_1,\ldots, v_t##, show ##u_1,\ldots,u_s## is linearly dependent
Relevant Equations:: Linear combination: c1u1+c2u2+c3u3+...+cnun=0
Linearly dependent is when there is a nontrival solution of the linear combination.

Solve for K1,K2,K3? or something else.
You do not need to solve for the Ks. What you need to do is to argue that a non-trivial solution exists.

I might have gone down another route, but yours should work fine too once generalised.

• PeroK
I am an typo. I gave the wrong inequality symbol. Sorry

It seems quite obvious that the problem statement is wrongly reproduced and should say ##s>t## instead of ##s<t##.

You do not need to solve for the Ks. What you need to do is to argue that a non-trivial solution exists.

I might have gone down another route, but yours should work fine too once generalised.
What is the other route? I might easier to understand.

I mean, it is in essence the same, but you get one constant less. Try to write ##u_s## as a linear combination of the first ##t## ##u_i##. (You can assume that the first ##t## ##u_i## are linearly independent because if they are not then you already know that the ##u_i## are linearly dependent.)

I mean, it is in essence the same, but you get one constant less. Try to write ##u_s## as a linear combination of the first ##t## ##u_i##. (You can assume that the first ##t## ##u_i## are linearly independent because if they are not then you already know that the ##u_i## are linearly dependent.)

If you can use some basic results or theorems about bases and dimensions of linear spaces, then the proof follows without further explicit calculation.

How can I do it without any theorems about bases and dimensions of linear spaces?

How can I do it without any theorems about bases and dimensions of linear spaces?
By explicit calculations that are similar to those used in the proofs of the relevant theorems.

• PhDeezNutz and FactChecker
How can I do it without any theorems about bases and dimensions of linear spaces?
Is this problem presented in a vacuum? Aren't there any relevant theorems that can be used and that they expect you to use?

Yes. My Monday lecture did not spoke about bases or dimensions. But we did take about linear combination and linearly independent sets as well as a vector space.

• PeroK