# Checking for linear independence of certain vectors

1. Mar 13, 2014

### negation

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

Given that { u1, u2, u3, u4, u5, u6 } are linearly independent vectors in R16, and that w is a vector in R16 such that w ∉ span{ u1, u2, u3, u4, u5, u6 }.

a) Is the set { 0, u1, u5 } is linearly independent?
b) the set { u1, u2, u3, u4, u5, u6, w } is linearly independent?
c) the set { u1, u4, u6 } is linearly independent ?

3. The attempt at a solution

$Span{u1, u2, u3, u4, u5, u6 }\in R16$

γ{ u1, u2, u3, u4, u5, u6 } = 0
(γ1u1 + γ2u2 + γ3u3 + γ4u4 + γ5u5 + γ6u6) = 0

w ∉ span{ u1, u2, u3, u4, u5, u6 } so that means there are no linear combinations for which
{ u1, u2, u3, u4, u5, u6 } spans w. How do I incorporate this into the approach?
If there are no linear combinations, then it stands to reason that the solutions to
{ u1, u2, u3, u4, u5, u6 } is inconsistent.

2. Mar 13, 2014

### Staff: Mentor

Start by answering the questions in order. It's not clear to me what you're doing.

Since the concept of linear independence is central to this problem, what is your definition of a set of linearly independent vectors? This concept is very subtle, and most beginning linear algebra students don't get it.

3. Mar 13, 2014

### negation

Let A be a set of vectors.

Span(A) = 0 is linearly independent iff λ1v1 + λ2v2 + . . . + λnvn = 0 and for all scalars λ, λ = 0

In essence, it must fulfill 2 sufficient condition for A set of vectors to qualify as being linearly independent:
1) the linear combination of A set of vectors λ1v1 + λ2v2 + . . . + λnvn = 0
If λ1v1 + λ2v2 + . . . + λnvn =/= 0 then it is linearly dependent.

2) for all scalar, λ, λ must be zero.
if at least 1 scalar, λ, is =/= 0, then, the set is linearly dependent.

It's 3am here but gotta press on. A couple more practice questions before I attempt the assessed questions.

Last edited: Mar 13, 2014
4. Mar 13, 2014

### Staff: Mentor

OK, I think you understand the concept of linear independence. The key idea is that for a set of lin. independent vectors, the equation λ1v1 + λ2v2 + ... + λnvn = 0 has exactly one solution; namely, λ1 = λ2 = ... = λn = 0.

Now to address what you wrote in post #1.
What does this mean?
This equation doesn't tell us anything. The six vectors could linearly independent or linearly dependent, all depending on how many solutions for the constants there are.
We don't talk about a set of vectors spanning another vector - you should say "there is no linear combination of <the six vectors> that equals w."
Start with the equation c1u1 + c2u2 + c3u3 + c4u4 + c5u5 + c6u6 + w = 0. How many solutions are there? (There is always at least one solution.)

Note that this is part b. Have you already done part a? What about part c?

5. Mar 13, 2014

### negation

You're right. My definition was lacking in rigor then.

I haven't attempted part (a)
Allow me to produce an attempt.

6. Mar 13, 2014

### negation

Part (a) :

span {0, u1, u5} = { u1, u2, u3, u4, u5, u6 }
λ1 (0, u1, u5) = { u1, u2, u3, u4, u5, u6 }
0λ1 + λ1u1 + λ1u5 = { u1, u2, u3, u4, u5, u6 }

I'm rather lost.

Edit: how do I write R16 as vectors?

Last edited: Mar 13, 2014
7. Mar 13, 2014

### LCKurtz

I don't think so. "Span(A)=0" is nonsense and the two statements are a confused mishmash of the required concepts. After watching several of negation's threads, my advice is that he badly needs to schedule some meeting time with his teacher.

Last edited: Mar 13, 2014
8. Mar 13, 2014

### Staff: Mentor

What I meant was "more or less understand the concept". I agree that Span(A) = 0 is nonsense.

9. Mar 13, 2014

### negation

I do think it's a tough unit but apparently, I'm not the only one struggling. We're given the theorems and it's up to us to apply it to practice problems during tut. I personally do think the whole unit is a rush. Quite a large percentage of student repeated this unit the 3rd time. Phew* I haven't repeated any and have been faring pretty well for the other units-D's
This, not to mention that there is a 40% drop out rate for this unit. Failure rates are pretty high too. The definitions used in this topic has been very vital and unfortunately, I'm still getting use to the idea and definition behind this topic although I'd say with the practice questions, I'm starting to see the bigger picture and getting a "feel". In any case, if I have to put in the necessary effort, I'd do so.