Checking for linear independence of certain vectors

[negation]

Homework Statement

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 ?

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.

 

[Mark44]

Homework Statement

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 ?

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.

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.

 

[negation]

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.

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 got to press on. A couple more practice questions before I attempt the assessed questions.

 

[Mark44]

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 + ... + λ_nv_n = 0 has exactly one solution; namely, λ₁ = λ₂ = ... = λ_n = 0.

Now to address what you wrote in post #1.

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

γ{ u1, u2, u3, u4, u5, u6 } = 0

What does this mean?

(γ1u1 + γ2u2 + γ3u3 + γ4u4 + γ5u5 + γ6u6) = 0

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.

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.

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."

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.

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?

 

[negation]

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 + ... + λ_nv_n = 0 has exactly one solution; namely, λ₁ = λ₂ = ... = λ_n = 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."

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

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

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

 

[negation]

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?

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?

 

[LCKurtz]

Since the concept of linear independence is central to this problem, what is your definition of a set of linearly independent vectors?

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.

.

OK, I think you understand the concept of linear independence.

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.

 

[Mark44]

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

 

[negation]

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.

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.