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

[cbarker1]

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

 

[BvU]

Let $s=3$ and $t=2$

This is strange. In the problem statement it says 'If $s<t$ ' ?

 

[fresh_42]

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.

 

[Orodruin]

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.

 

[cbarker1]

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

 

[cbarker1]

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.

 

[Orodruin]

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.)

 

[cbarker1]

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.)

Proof by contradiction, right?

 

[PeroK]

Proof by contradiction, right?

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

 

[cbarker1]

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

 

[PeroK]

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.

 

[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?

 

[cbarker1]

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.