Question about polynomial vector space

[cotfw]

I don't understand how Pn (the set of polynomials whose degree is equal to or below n, and n >= 0) can be a vector space, because it doesn't seem to be closed under multiplication. For example, if I take P2, and I have this basis {1,x,x²}, and I create this linear combination of them:

1 + x + x²,

and then I multiply it by x, then it will become:

1x + x² + x³

which is in P3, not P2. I am having trouble understanding this.

 

[Steven G]

I don't understand how Pn (the set of polynomials whose degree is equal to or below n, and n >= 0) can be a vector space, because it doesn't seem to be closed under multiplication. For example, if I take P2, and I have this basis {1,x,x²}, and I create this linear combination of them:

1 + x + x²,

and then I multiply it by x, then it will become:

1x + x² + x³

which is in P3, not P2. I am having trouble understanding this.

I agree that if you multiply 1 + x + x² by x then you are no longer in P2. The issue here is that in a vector space you can multiply only by a scalar--like 7 or pi or 11/2.

Think about this. (1,3, 7) and (-3, 5, 90) are vectors in R^3. can you multiply them? (No!). can you multiply each of those vectors by 6? (Yes).

Now go back and read the definition of a vs a few times as it is very important that you know this definition.

BTW 1x + x² + x³ is not necessarily in P3. It can be in P4 or P5 or in Pn, where n is any integer greater than 2

 

[cotfw]

But I don't understand why x can't be a scalar? X is just a stand in for a scalar, isn't it?

 

[Steven G]

But I don't understand why x can't be a scalar? X is just a stand in for a scalar, isn't it?

The technical difference between a vector and scalar is that a scalar is one dimensional (like height, weight, temperature,...) and a vector is..well a vector. It has a magnitude and it has a direction.

In a vector space you can multiply a vector just by a scalar and this scalar is for the most part not in the vector space. Again, consider the vector space R^3. (1,2,3) is in R^3. If we think of the vector (4,5,6) in R^3 as a scalar, then how do we multiply these? We can however multiply (1,2,3) by 5 and we get 5(1,2,3) = (5, 10, 15) which is a vector in R^3.

This multiplication of x you speak of can be thought of as (0,1,0) in R^3 and therefore has a magnitude and direction.

 

[Ishuda]

But I don't understand why x can't be a scalar? X is just a stand in for a scalar, isn't it?

Or, to put it another way which may help, think of a vector space as n parts (dimensions, spaces), (a, b, c) for example when there are three parts or, if there is just 1 dimension part (a). It has the parens around it to indicate it is a vector [other notations could also be used such a bold face type or a bracket or ...]. Thus the 3 space vector x would be designated as (a, b, c) which could (and in this example does) mean a + b x + c x² where a, b, c, and x are scalars. So, although you could use just a plain x to mean both a scalar and the vector (0, 1, 0), you still have to keep them separate when you start to manipulate them, i.e. add. multiple, etc.

 

[cotfw]

Would it be possible to explain it without referring to R3? I can understand why R3 s a vector space, but P3 is much more confusing.

I know the difference between a vector and a scalar, but I just can't understand why the x in x(x³) would be considered a vector. A vector has a direction and a magnitude, so where is the direction in this case?

 

[Steven G]

Would it be possible to explain it without referring to R3? not that I can think of
I can understand why R3 s a vector space, but P3 is much more confusing. it should not be if you accept that R^3 and P2 are the same exact set with different labels.

I know the difference between a vector and a scalar, but I just can't understand why the x in x(x³) would be considered a vector. A vector has a direction and a magnitude, so where is the direction in this case?

Consider the mapping F2--> R^3 defined as follows. F(ax^2+bX+c) = (a,b,c)

Let A=ax^2+bx+c and B=a'x^2+b'x+c be in P2.

Suppose F(A) = F(B). Then (a,b,c)=(a',b',c'). So a=a', b=b', c=c'. ....so A=B and F is 1-1

F is onto since if (r,s,t) is in R^3 then F (rx^2 + sx + t) = (r,s,t)

It is easy to show that F(A+B) = F(A) + F(B)

Let r be any real. Please verify that F(rA) = rF(A)

Then we established that R^3 is isomorphic to P2

This is the part that you need to get comfortable with: This means that P2 and R^3 are the same vector space just with the vectors in P2 named differently then the vectors in R^3.

Since every vector in R^3 has a direction and magnitude then the corresponding vector in P2 also have the same magnitude and the same direction.

I hope this helps

 

[Steven G]

A vector has a direction and a magnitude, so where is the direction in this case?

Lets say that v = 3x^2 + 4x - 2 is in P2. Then the magnitude of v is sqrt(3^2 + 4*2 + (-2)^2) and is in the direction of <3,4,-2>

 

[HallsofIvy]

I don't know whether to complain or just bite my tongue at Jomo's "it has a magnitude and it has a direction." We are not talking about Physics vectors in space here, we are talking about abstract Linear Algebra. And neither "magnitude" nor "direction" are defined in a general vector space.

However, any n- dimensional vector space is isomorphic to \(\displaystyle R^n\) so we can define "inner product", then "magnitude" and "direction" but you should understand that there are many such isomorphism (one for every choice of basis for the vector space) and different isomorphism may give a different "magnitude" and "direction" for the same vector.

The answer to cotfw's original complaint, "I don't understand how Pn can be a vector space, because it doesn't seem to be closed under multiplication", which has already been given, is that only two operations are defined for a vector space- addition of vectors and multiplication of a vector by a number (scalar). Multiplication of vectors is not, in general, defined for a vector space.

Of course, we can define multiplication of polynomials so we have a little more structure than just a vector space. An "algebra" is defined to be a vectors space in which multiplication of vectors is defined in such a way that certain properties (such as the associative property) are obeyed and the vector space is closed under that multiplication. You are correct that Pn, the set of all polynomials of degree n or less, is NOT closed under multiplication and so not an "algebra". The set of all polynomials, of any degree, is closed under multiplication and so is an "algebra".