Checking for linear independence of certain vectors

In summary: + 0 = (0, 0, 0)λ1 (0, 0, 0)+ λ2 (u1, u2, u3)+ λ3 (0, 0, 0)λ2 (u1, u2, u3) = 0λ2 (0, 0, 0)+ λ3 (0, u1, u5)λ3 (0, u1, u5) = 0λ3 (0, 0, 0)+ λ4 (u1, u2, u3)+ λ5 (u4, u5, u6)λ4 (u1, u2, u3) = 0
  • #1
negation
818
0

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




[itex]Span{u1, u2, u3, u4, u5, u6 }\in R16[/itex]

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.
 
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  • #2
negation said:

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




[itex]Span{u1, u2, u3, u4, u5, u6 }\in R16[/itex]

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.
 
  • #3
Mark44 said:
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.
 
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  • #4
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.
negation said:
[itex]Span{u1, u2, u3, u4, u5, u6 }\in R16[/itex]

γ{ u1, u2, u3, u4, u5, u6 } = 0
What does this mean?
negation said:
(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.
negation said:
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."
negation said:
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?
 
  • #5
Mark44 said:
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."

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

Mark44 said:
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.
 
  • #6
Mark44 said:
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?
 
Last edited:
  • #7
Mark44 said:
Since the concept of linear independence is central to this problem, what is your definition of a set of linearly independent vectors?

negation said:
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.

.

Mark44 said:
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.
 
Last edited:
  • #8
What I meant was "more or less understand the concept". I agree that Span(A) = 0 is nonsense.
 
  • #9
LCKurtz said:
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.
 

1. What is linear independence of vectors?

Linear independence refers to a set of vectors in a vector space that cannot be represented as a linear combination of the other vectors in the set. In other words, none of the vectors in the set can be derived from a combination of the other vectors.

2. How do you check for linear independence of vectors?

To check for linear independence of vectors, you can use the row reduction method, also known as Gaussian elimination. This involves creating a matrix with the vectors as rows, and then performing row operations to determine if the rows can be reduced to a row of zeros. If they cannot, the vectors are linearly independent.

3. Why is it important to check for linear independence of vectors?

Checking for linear independence of vectors is important because linearly dependent vectors can lead to redundancy and inefficiency in mathematical calculations. By ensuring linear independence, we can accurately represent and manipulate data or systems using the vectors.

4. Can vectors be linearly independent in one vector space and linearly dependent in another?

Yes, it is possible for vectors to be linearly independent in one vector space and linearly dependent in another. This is because the definition of linear independence depends on the specific vector space and the operations that can be performed within that space.

5. Are there any other methods for checking linear independence of vectors besides row reduction?

Yes, there are other methods such as the determinant method, the rank method, and the Gram-Schmidt process. Each method has its own advantages and may be more suitable for certain types of vector sets. It is always beneficial to have multiple methods for checking linear independence in order to confirm the results and find the most efficient approach.

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