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

• cbarker1
In summary: 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.
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

cbarker1 said:
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##.

cbarker1 said:
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

Orodruin said:
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.)

Orodruin said:
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 said:
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?

cbarker1 said:
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
cbarker1 said:
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

## 1. What does it mean for one set v to be a linear combination of u?

A set v is said to be a linear combination of another set u if every element in v can be expressed as a linear combination of the elements in u. This means that each element in v can be written as a scalar multiple of an element in u, with all the scalar multiples added together.

## 2. How do you prove that u is linearly dependent?

To prove that u is linearly dependent, you need to show that there exists a non-zero solution to the equation c1u1 + c2u2 + ... + cnun = 0, where c1, c2, ..., cn are scalars and u1, u2, ..., un are the elements in u. This means that there is a non-trivial way to combine the elements in u to get a sum of zero.

## 3. What is the difference between linearly dependent and linearly independent sets?

A set is linearly dependent if there exists a non-trivial way to combine its elements to get a sum of zero. On the other hand, a set is linearly independent if there is no non-trivial way to combine its elements to get a sum of zero. In other words, the only way to get a sum of zero is by multiplying all the elements by zero.

## 4. Can you give an example of a linearly dependent set?

One example of a linearly dependent set is {v1, v2, v3} where v1 = (1, 2, 3), v2 = (2, 4, 6), and v3 = (3, 6, 9). In this set, v3 is a linear combination of v1 and v2, since v3 = 3v1 + 3v2. Therefore, this set is linearly dependent.

## 5. How is the linear dependence of a set related to its span?

The linear dependence of a set is closely related to its span. If a set is linearly dependent, it means that at least one of its elements can be written as a linear combination of the other elements. This means that the span of the set is not equal to the number of elements in the set. On the other hand, if a set is linearly independent, its span is equal to the number of elements in the set.

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