Resolving Vector in Arbitrary Basis: Proving Coefficients' Independence

In summary: I think now, please correct me if I'm wrong:1. a, b, and c form a basis, which implies that they are independent and span the space.2. v can be written as a linear combination of a, b, and c, so v lies in the span of a, b, and c.3. If α, β, and γ are the coefficients of the linear combination, then v = αa + βb + γc.4. Since v lies in the span of a, b, and c, it can be written as a linear combination of these basis vectors.5. This means that there is no overlap between v and either b or c, because if there were
  • #1
precise
13
0
Hi, I came across this statement while I was reading an article on resolution of vector in arbitrary basis.

"v = αa + βb + γc ---(1) , where a, b and c are three independent vectors.
we observe that the coefficient α cannot involve any overlap of v with either b or c ; β cannot
involve any overlap of v with either c or a ; and γ cannot involve any overlap of v with either
1a or b. Therefore α must be proportional to that part of v which lies along ( b × c ), i.e., to
[ (b × c) · v ]. Similar conclusions hold good for β and γ."

This is a part of the proof that says v = [a [ (b × c) · v ] + b [ (c × a) · v ] + c [ (a × b) · v ]]/[ (a × b) · c ]

My question is, how can we ascertain that " the coefficient α cannot involve any overlap of v with either b or c" from the given equation (1)
Here is the link to what I was reading in page 2,
http://www.ias.ac.in/resonance/Volumes/01/10/0006-0013.pdf
 
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  • #2
My question is, how can we ascertain that " the coefficient α cannot involve any overlap of v with either b or c" from the given equation (1)
It is what "resolving a vector in a basis" means.
 
  • #3
That is not a proof, that is a classical example of hand waving. "Coefficient involve any overlap", hello? That can mean just about anything the author wants that to mean. Unless there is a clear definition of the meaning of that, it cannot be ascertained.

I am not entirely satisfied with Simon Bridge's response, either, because it assumes that the procedure of "resolving a vector in a basis" has been unambiguously specified. This is very frequently not the case; the procedure is typically given for orthogonal bases, and the non-orthogonal bases are brushed aside by saying that for any non-degenerate basis there is one and only one set of coefficients that represent any vector (as follows from a theorem in linear algebra). And it seems to me that the "proof" is actually an attempt to specify the procedure, so certainly it cannot be assumed as already given.
 
  • #4
I am not entirely satisfied with Simon Bridge's response, either, because it assumes that the procedure of "resolving a vector in a basis" has been unambiguously specified. This is very frequently not the case...
... that is sadly so often the case.

Unfortunately we do not have the rest of the text or the lesson or the context... I am hoping that OPs response will shed light on that.
 
  • #5
As far as I understand, when we try to resolve v in the basis of a b and c according to equation (1), the coefficient α, β, or γ are scalar values, which means they must be the resulting from the dot product of vectors. What's wrong in assuming that α can contain the dot product among any two of the vectors v, a, b and c, like v.a, v.b or b.c, etc. Is there any reasoning behind this?

Here is the link to what I was reading in page 2,
http://www.ias.ac.in/resonance/Volumes/01/10/0006-0013.pdf
 
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  • #6
The point is to find α, β, and γ that satisfy v = αa + βb + γc. Take the dot product.
(b × c) · v = (b × c) · (αa + βb + γc) = α[(b × c) · a] + β[(b × c) · b] + γ[(b × c) · c] = α[(b × c) · a] + 0 + 0, hence α = [(b × c) · v] / [(b × c) · a], and similarly for β and γ.
 
  • #7
dauto said:
The point is to find α, β, and γ that satisfy v = αa + βb + γc. Take the dot product.
(b × c) · v = (b × c) · (αa + βb + γc) = α[(b × c) · a] + β[(b × c) · b] + γ[(b × c) · c] = α[(b × c) · a] + 0 + 0, hence α = [(b × c) · v] / [(b × c) · a], and similarly for β and γ.
Hi Dauto, I thought of that, it's clear. But I'm trying to understand the author's intuition behind the statement "the coefficient α cannot involve any overlap of v with either b or c" starting with the equation v = α a + β b + γ c ".
 
  • #8
precise said:
coefficient
Hello,
I am just curious to know why you are using this "ffi" instead of "ffi" ? :smile:
 
  • #9
adjacent said:
Hello,
I am just curious to know why you are using this "ffi" instead of "ffi" ? :smile:

Didn't notice that. I copied the text from a page. It turned out like this.
 
  • #10
precise;4768597. said:
But I'm trying to understand the author's intuition behind the statement "the coefficient α cannot involve any overlap of v with either b or c" starting with the equation v = α a + β b + γ c ".
The reason you are finding it difficult to understand is because it is a very peculiar thing to write. I guess the author is alluding to the fact a,b and c form the basis vectors of a 3 dimensional space so, loosely speaking, form the 'building blocks' or 'directions' for the construction of all vectors in said space.
 
  • #11
precise said:
Hi Dauto, I thought of that, it's clear. But I'm trying to understand the author's intuition behind the statement "the coefficient α cannot involve any overlap of v with either b or c" starting with the equation v = α a + β b + γ c ".

I can't help you with that. I think that statement makes it seem harder than it really is.
 
  • #12
CAF123 said:
I guess the author is alluding to the fact a,b and c form the basis vectors of a 3 dimensional space so, loosely speaking, form the 'building blocks' or 'directions' for the construction of all vectors in said space.

OK, let's assume that a, b and c are basis vectors in 3D space. But how does that lead to the conclusion "the coefficient α cannot involve any overlap of v with either b or c" starting with the equation v = α a + β b + γ c ".
 
  • #13
Simon Bridge said:
... that is sadly so often the case.

Unfortunately we do not have the rest of the text or the lesson or the context... I am hoping that OPs response will shed light on that.

Please check, updated the question with source of reference.
 
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  • #14
precise said:
Hi Dauto, I thought of that, it's clear. But I'm trying to understand the author's intuition behind the statement "the coefficient α cannot involve any overlap of v with either b or c" starting with the equation v = α a + β b + γ c ".

Footnote 1 makes that clear. So, not so hand wavy as it seemed originally.
 
  • #15
I'm trying to understand the author's intuition ...
The author is just trying to be nice to the student by describing things in terms of concepts he thinks the student should find familiar. In your case that was not successful.

Don't worry about it - just try to learn the maths and your own intuition will develop.

In 2D terms, for simplicity, I am guessing you are thinking something like:
To describe a 2D space, you can use any two vectors that do not point in the same direction.
If the angle between them is not pi/2 then any movement in the direction of one will involve a movement in the direction of the other. i.e. you can resolve one of the "basis vectors" against the other one.

But, according to the footnote, that is not what is meant by "overlap".
All he's really saying is that the vector is uniquely specified by the coefficients of the basis vectors, in that basis.

The simple answer to your original question is that you cannot get there from equation (1) alone, and the author does not either.
 

Related to Resolving Vector in Arbitrary Basis: Proving Coefficients' Independence

1. What is the purpose of resolving vectors in an arbitrary basis?

The purpose of resolving vectors in an arbitrary basis is to express the components of a vector in terms of a different set of basis vectors. This allows for easier calculations and analysis, especially when dealing with vectors in non-standard coordinate systems.

2. How is the independence of coefficients in resolving a vector proven?

The independence of coefficients in resolving a vector is proven by using the Gram-Schmidt process. This process involves finding an orthogonal basis for the vector space and then normalizing the basis vectors to ensure that they have unit length. The resulting coefficients will be independent of each other.

3. Can resolving vectors in an arbitrary basis be applied to any vector?

Yes, resolving vectors in an arbitrary basis can be applied to any vector as long as the basis vectors are linearly independent. This means that they cannot be expressed as a linear combination of each other.

4. What is the difference between resolving vectors in an arbitrary basis and standard basis?

The standard basis for a vector space is a set of basis vectors that are perpendicular to each other and have unit length. Resolving vectors in an arbitrary basis involves finding a new set of basis vectors that may not be perpendicular or have unit length. This allows for more flexibility in expressing vectors in non-standard coordinate systems.

5. How is resolving vectors in an arbitrary basis used in real-life applications?

Resolving vectors in an arbitrary basis is used in many real-life applications, such as computer graphics, engineering, and physics. In computer graphics, it is used to rotate and transform objects in a 3D space. In engineering, it is used to calculate forces and stresses in non-standard coordinate systems. In physics, it is used to analyze vector quantities in different reference frames.

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