# Homework Help: Question about linear independence

1. Mar 3, 2017

### Arnoldjavs3

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
if there exists a set with 3 vectors, and all of them are linear independent, then by definition no linear combination of the 3 vectors can equal to 0.

I believe that's an accurate definition right? So in this case, the answer is that it will always be linear independent regardless of what the value of every vector is, because they have already stated that the set is linear independent. The second part just shows an example of linear combinations correct?

2. Mar 3, 2017

### ehild

No. The second set is 4 vectors in a three-dimensional space.

3. Mar 3, 2017

### PeroK

I believe I could make a case that those vectors are always linearly dependent.

4. Mar 3, 2017

### PeroK

$u, v, w$ could be vectors in any n-dimensional space.

5. Mar 3, 2017

### ehild

But the space spanned by them is three dimensional.

6. Mar 3, 2017

### Arnoldjavs3

So if you could make a combination that indicates that they are dependent then it would mean that it depends on values of the vectors? I'm having a hard time putting this all in to perspective without examples.

How do you know this? I thought they were just 3, individually independent vectors in n dimensional space?

7. Mar 3, 2017

### PeroK

Okay, I see what you mean.
Suppose you take the two vectors: $u - v - w$ and $-2u + 2v + 2w$. Are they linearly independent?

8. Mar 3, 2017

### ehild

u, v, and w are linearly independent vectors. Adding any linear combination of them makes the set of the four vectors dependent.
What is the definition that a set of vectors are independent?

In n dimensional space there are n linearly independent vectors.

9. Mar 3, 2017

### Arnoldjavs3

No, because they are multiples of one another
So, in hte first set they are linear independent. But because set two has all vectors adding several multiples of each member it is linear dependent(hope this makes sense)? I didn't see that the second set had four members and not three. This makes alot more sense now

10. Mar 3, 2017

### ehild

The vectors in the second set are all linear combinations of vectors $\vec u$ , $\vec v$, $\vec w$. They are
$\vec a=\vec u+2\vec v +2\vec w$
$\vec b=2\vec u+5\vec v +2\vec w$
$\vec c=-3\vec u+2\vec v -4\vec w$
$\vec d=-\vec u+5\vec v +4\vec w$

11. Mar 4, 2017

### Staff: Mentor

I don't think you understand the definition of linear independence. Given any set of three vectors that belong to some vector space, the equation $c_1\vec{x_1} + c_1\vec{x_1} + c_1\vec{x_1} = \vec{0}$ always has a solution in terms of the constants $c_1, c_2, c_3$. This is true whether the vectors are linearly independent or linearly dependent.

The key difference is that for a set of linearly independent vectors, the only solution is $c_1 = c_2 = c_3 = 0$. For a set of linearly dependent vectors, there will be an infinite number of solutions, including $c_1 = c_2 = c_3 = 0$.

Last edited: Mar 5, 2017
12. Mar 4, 2017

### PeroK

But, your original argument was that if the vectors were linearly independent then no linear combination of them is 0, so no linear combination of linear combinations can be 0. But, as you see a linear combination of linear combinations can be 0. This is because a linear combination of linear combinations can end up having a 0 coefficient for each vector.

The simplest solution is that put forward by @ehild, as there are four vectors. But, what if you had only three vectors? Just the first three, say:

$u + 2v + 2w, 2u + 5v, -3u +2v -4w$

How could you tell whether these are linearly independent?