Struggling to understand vecot coordinate mapping.

[bobby2k]

I am somewhat puzzled after reading that polynomials can be vectors, this concept confuses me.

For instance, they can say that a basis for polynomials [itex]P_2[/itex] can be.
[itex]B=\{1+t^{2},t+t^{2},1+2t+t^{2}\}[/itex]

In this case will the mapping
[itex][1+t^{2}]_{B}[/itex]
be [1,0,0] or [1,0,1]?

 

[LCKurtz]

I am somewhat puzzled after reading that polynomials can be vectors, this concept confuses me.

For instance, they can say that a basis for polynomials [itex]P_2[/itex] can be.
[itex]B=\{1+t^{2},t+t^{2},1+2t+t^{2}\}[/itex]

In this case will the mapping
[itex][1+t^{2}]_{B}[/itex]
be [1,0,0] or [1,0,1]?

##1+t^2 = 1\cdot(1+t^2) + 0\cdot(t+t^2)+ 0\cdot(1+2t+t^2)##. Is that what you are asking?

 

[Mark44]

I am somewhat puzzled after reading that polynomials can be vectors, this concept confuses me.

The polynomials aren't vectors, but there is pairing between specific function spaces (such as P₂, the space of polynomials of degree 2 or less) and the corresponding vector spaces (such as R³).

For instance, they can say that a basis for polynomials [itex]P_2[/itex] can be.
[itex]B=\{1+t^{2},t+t^{2},1+2t+t^{2}\}[/itex]

In this case will the mapping
[itex][1+t^{2}]_{B}[/itex]
be [1,0,0] or [1,0,1]?

How would you write 1 + t² as a linear combination of the functions in your basis B?

 

[bobby2k]

The polynomials aren't vectors, but there is pairing between specific function spaces (such as P₂, the space of polynomials of degree 2 or less) and the corresponding vector spaces (such as R³).

How would you write 1 + t² as a linear combination of the functions in your basis B?

Yes I see now that the first one is correct, thanks.

I just have one more question. A problem arises in my mind when they allow polynomials and vectors to be connected.

First: Wikipedia defines the dimension of a vector space to be:
"In mathematics, the dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis of V."

Now look at this proof from my text-book.

The thing that I do not understand is how can they know how many entries there are in the baisis-vectors? If we now assume that there is the pairing that you mentioned, how can we be sure that the coordinate vectors have n entries? The reason I find this weird is because we are working with polynomials, or some other abstract thing. And we say that the objects {b1,b2,..bn} is a basis if they are linerly independent and every object that they are supposed to span can be written as a linear combination of them, correct? But, when we define it this way we are not working with the traditional vectors at all, they are not mentioned! All we know is that we have n objects.

 

[voko]

The thing that I do not understand is how can they know how many entries there are in the baisis-vectors?

The say in the beginning "let the space have a basis with n vectors". Then there must be just n "entries" - because each entry corresponds to one basis vector.

But, when we define it this way we are not working with the traditional vectors at all, they are not mentioned! All we know is that we have n objects.

Traditionally, vectors are just things with "n entries". When n = 2 or n = 3, there is also a geometric interpretation that associates them with "arrows".

 

[LCKurtz]

Yes I see now that the first one is correct, thanks.

I just have one more question. A problem arises in my mind when they allow polynomials and vectors to be connected.

First: Wikipedia defines the dimension of a vector space to be:
"In mathematics, the dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis of V."

Now look at this proof from my text-book.

The thing that I do not understand is how can they know how many entries there are in the baisis-vectors? If we now assume that there is the pairing that you mentioned, how can we be sure that the coordinate vectors have n entries? The reason I find this weird is because we are working with polynomials, or some other abstract thing. And we say that the objects {b1,b2,..bn} is a basis if they are linerly independent and every object that they are supposed to span can be written as a linear combination of them, correct? But, when we define it this way we are not working with the traditional vectors at all, they are not mentioned! All we know is that we have n objects.

The say in the beginning "let the space have a basis with n vectors". Then there must be just n "entries" - because each entry corresponds to one basis vector.


Traditionally, vectors are just things with "n entries". When n = 2 or n = 3, there is also a geometric interpretation that associates them with "arrows".

That's true if your vector space is ##\mathcal R^n## but not for general vector spaces. If your vector space is a space of functions, you can't talk about "components" and geometric representations like "arrows". Bobby2k pretty much has it correct about that. The argument in the theorem he quoted is for ##\mathcal R^n##.

 

[voko]

That's true if your vector space is ##\mathcal R^n## but not for general vector spaces. If your vector space is a space of functions, you can't talk about "components" and geometric representations like "arrows". Bobby2k pretty much has it correct about that. The argument in the theorem he quoted is for ##\mathcal R^n##.

I think you misunderstood. My intent was to hint at that even traditionally vectors are much more than "arrows". Polynomials (up to a fixed degree) fit the bill, so there is no reason to go beyond traditional just yet.

 

[bobby2k]

I think I understand what you mean now. So linear algebra can be used for many different things I guess.
The reason polynomials can be used in linear algebra is because if we add two polynomials then the resulting polynomial is the one we get if we add each part? So linear algebra can be used on all objects which have the property that if you add them, then each individual property is added together, and if you scale the object then each property is scaled.

Is this abstract definition valid?: If we look at a pencil, then this pencil is a one-dimensional basis, and we say that the property we are looking at is how much we can write with that pencil. Then if we add two pencils(of this type) of equal lengths we can write double as much, and if we define scaling of a pencil as the scaling of the length, than this will also be proportional to the time we can write?

(But a woman is not a basis for a vector space if you look at the property of complaining. Because if you add two women the resulting effect will be more complaining than the sum of each individuals womans complaining?:D)

 

[voko]

So linear algebra can be used on all objects which have the property that if you add them, then each individual property is added together, and if you scale the object then each property is scaled.

Precisely! That's what defines a 'vector'.

 

[1MileCrash]

So linear algebra can be used on all objects which have the property that if you add them, then each individual property is added together, and if you scale the object then each property is scaled.

That's rather enlightening.

I'm studying linear algebra too right now, and the way I understand it is that if a set has the properties of being a vector space (my book names 8 basic properties) then any element of that set is a vector, no matter what those elements may be (like polynomials). The things you just said, I think, are direct corollaries to those properties. My book even says "the reader should not confuse this use of the word "vector" with the physical entity discussed in section 1.1; the word "vector" is now being used to describe any element of a vector space."

Fortunately I already had to get used to the idea that a vector isn't really just a pointy arrow (by that I mean it can be generalized) in tensor analysis.

In tensor analysis we defined a vector in N-space as an N-tuple that followed a certain transformation law. I wonder if that transformation law and properties inherent from being a member of a vector space are equivalent. Hmm.

 

[bobby2k]

In tensor analysis we defined a vector in N-space as an N-tuple that followed a certain transformation law. I wonder if that transformation law and properties inherent from being a member of a vector space are equivalent. Hmm.

This was very interesting. I haven't worked with tensor myself, but I have had some mechanics. I remember if you had a very complex statically indeterminate system, you could build a big stifness matrix and solve the problem with linear algebra.

There are two properties I remember, which I allways thought were "physical" properties, but now I think may be mathematical.

1. The superposition principle: The principle that you can look at the effect of forces seperately and then add the movement caused by each force together.
2. Hookes law. That strain is proportional to stress. This translates to movement beeing proportional to the forces.

These two properties are "physical". But if you look at them carefully, they are exactly the same properties that is used to define a subspace in linear algebra. So this begs the question: Where these properties made because they really fit the physical properties(for small deflections etc.), or where they primarily made to take the physical problem into the mathematical world?, and the vector space?

 

[voko]

1. The superposition principle: The principle that you can look at the effect of forces seperately and then add the movement caused by each force together.
2. Hookes law. That strain is proportional to stress. This translates to movement beeing proportional to the forces.

These two properties are "physical". But if you look at them carefully, they are exactly the same properties that is used to define a subspace in linear algebra. So this begs the question: Where these properties made because they really fit the physical properties(for small deflections etc.), or where they primarily made to take the physical problem into the mathematical world?, and the vector space?

Re (1), recall Newton's second law - in vector form. What do you think?

(2) is a tad trickier. Hooke's law, as you know, is not exact: it works only when the stress is small. And then we have the apparatus of differential calculus, which allows us to approximate arbitrary functions with linear functions when changes in argument are small. Do you see the connection with vectors here?