Imagine you are given 5 polynomiums. How do you decide which of them are a basis of a vector space?
Does anybody know how I should start? Is it possible for just 1 polynomium to be a basis of a space? I don't think so, so that means I need to pick at least two of the 5 polynomiums, correct?
your question is too vague to give a really good answer. a basis has to be linearly independent, and span the space you're checking to see if it's a basis OF.
to check for spanning, you try to show that any polynomial in the space you want a basis for, is a linear combination of elements of the set you're hoping might be a basis.
to check for linear independence, you show that if a linear combination of your set elements is 0, all the coefficents in the linear combination must also be 0.
if you are given a set of 5 polynomials, {p1,p2,p3,p4,p5} and want to know if they are a basis for span({p1,p2,p3,p4,p5}), you don't have to check for spanning, just linear independence. but if you want to show they are a basis for "all polynomials of degree k or less" you DO have to check spanning.
it IS possible to have a basis with a single element. for example, {1} is a basis for the space of all "constant polynomials" (polynomials of degree 0).
What if let's say I'm given 5 polynomiums:Originally Posted by Deveno
How do I pick a basis for a subspace to the space
if I have to choose amongst those 5 polynomiums?
The subspace is
Do I have to test out every single combination of the 5 polynomiums to see which is a linear combination? That is a really tedious job to do and I don't think that is the correct way to go about with this. Have I missed something?
I have tried to put each of the polynomiums in a 3x3 matrix (with two 0 rows) and then calculate the determinant but all of the determinants equal 0 so I don't think that was the correct method.
that is an entirely different, and much more specific question.
P2(R) has dimension 3 (one basis is {1,x,x^2}) so any OTHER basis for it must also have 3 elements. on the other hand, if you find 3 linearly independent elements of P2(R), you automatically have a basis.
if you want to find a basis for the subspace {p(x) : p(2) = 0}, first you have to determine ITs dimension. first of all, note that the only constant polynomial p(x) = c that satisfies p(2) = 0 is the 0-polynomial. that means our subspace can have dimension at MOST 2 (an entire dimension of P2(R) is not in our subspace). on the other hand, it has to have dimension at least 2, because x-2 and x^2 - 3x + 2 are both in it, and if:
a(x - 2) + b(x^2 - 3x + 2) = 0 <-- 0-polynomial, not the number
ax - 2a + bx^2 - 3bx + 2b = 0
bx^2 + (a-3b)x + (a+2b) = 0 so
b = 0
a-3b = 0
a+2b = 0, so a = b = 0, these two polynomials are linearly independent. so the dimension of your subspace is 2.
the first thing you must do, is determine which of your 5 polynomials are actually IN the subspace. you will be able to eliminate some of them. then, if you have more than 2 left, pick one. that will be the start of your basis. then pick another one, and test for linear independence. if 2 pass the linear independence test, you have a basis.
Aha, I see.
But I'm a bit confused about x-2 and x^2 - 3x + 2. Where did you get these from?
And doesn't P(2) mean that x=2? So the x should be substituted with 2?
well x-2 is obvious? suppose p(2) = 0. this means that either:
1) p is the 0-polynomial (which is 0 for every x)
2) p contains the factor (x-2):
if p(x) = (x-2)g(x), then p(2) = (2-2)g(2) = 0g(2) = 0, no matter "what" g(x) is.
now suppose p(2) ≠ 0. we can write p(x) = (x-2)q(x) + r(x), by dividing p(x) by x-2. now p(2) = (2-2)q(2) + r(2) = r(2), and p(2) ≠ 0, so r(2) ≠ 0, so in particular, r(x) is NOT the 0-polynomial.
but this means that (x-2) is not a factor of p(x), we get a "non-zero" remainder.
so where did i get x^2 - 3x + 2 from? i just made up a polynomial of the form (x - 2)(x + a), in this case, i picked a = 1, x^2 - 3x + 2 = (x-2)(x+1).
you can verify directly that (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0, or you can see that (2-2)(2+1) = (0)(3) = 0, both ways show that x^2 - 3x + 2 is in the subspace {p(x) : p(2) = 0}.
in other words, another way of characterizing our subspace is U = {p(x) : p(x) = (x-2)g(x)}.
now, out of your 5 polynomials, which ones have x-2 as a factor?
SinceWhat if let's say I'm given 5 polynomiums:
How do I pick a basis for a subspace to the space
if I have to choose amongst those 5 polynomiums?
The subspace is
Do I have to test out every single combination of the 5 polynomiums to see which is a linear combination? That is a really tedious job to do and I don't think that is the correct way to go about with this. Have I missed something?
I have tried to put each of the polynomiums in a 3x3 matrix (with two 0 rows) and then calculate the determinant but all of the determinants equal 0 so I don't think that was the correct method.is a basis for this space you need three polynomials so that you can form each of the known basis by a linear combination of them. It is obvious the first three of your polynomials will do, so now provide the demonstration.
CB
Shouldn't (x-2)(x+1) be equal to x^2-x-2?
I mean, the expression is x*x=x^2 and then x*1=x and -2*x=-2x so x-2x=-x
And -2*1=-2
That would still give us zero: (2)^2 -1*2 - 2 = 0
So to find which polynomiums that can be accepted:
(x-2)(1-2)=x-2x-2+4=-x+2
So the first polynomium (2-x) is a basis for the subspace.
(2-x^2) cannot be factorized by (x-2) so this is not a basis.
(4-3x) cannot be factorized either.
(20-9x+x^2) cannot be factorized either.
(12-8x-x^2) cannot be factorized either.
So that means only 1 of them is a basis in the subspace?
If I had found two of the polynomiums that were basis for the subspace, then I could do a test for linear dependance:
p1 + p2 = 0
If this statement for my two polynomiums turns out to be true then something went wrong since a basis always have to be linear independant.
But if that statement turns falsethen I have my two basis for the subspace.
Is that correct?
x - 2 = 1(x -2), every polynomial is a factor of itself (just like every integer divides itself, it's the same principle). you might want to re-check the polynomials as they are listed. in your first post, you gave:
, which is actually
. if you don't have two polynomials in your list of 5, you won't be able to have a basis (you need 2 dimensions, but a single basis element just gives you 1).
happens to me more than i'd like to admit....
Okay, so let me see if I have understood this correctly.
In order to find out if 5 polynomials are basis for the subspace, I need to first find the root for the subspace.
Since P(2)=0 it means that x=2, so in order for the expression to equal 0, we need the factor (x-2) since 2-2=0
Since this is the root, we can check it with the polynomials and see which of the polynomials that has this factor as a root.
If the polynomial can be expressed as a factor with (x-2) then it is in our subspace.
Once we have chosen at least 2 of the polynomials, we can see that these polynomials are a basis for our subspace.
Am I understanding this correctly?
they will only be a basis IF they are linearly dependent AND span.
but...as we saw before, our subspace has dimension 2 (we found a basis with 2 elements, and we know the dimension is not 3). so you can get by with just checking for linear independence (you should understand why this is true, and it seems like you may not).
but they HAVE to be linearly independent, our else you don't get "the whole subspace". for example, x-2 and 2x-4 wouldn't work.
Are you sure it's linearly dependent? Because in my textbook it says that the vectors in a vectorset are basis for the space if they span the space and if they are linearly independent.
So I have to check if the vectors are linearly independant and if they are, they are a basis of my space?
Our subspace has the dimension 2 because there was 2 polynomials that could be factorized with (x-2).
Our space has the dimension 3 because the polynomial P_2(R) has the basis (1,x,x^2) which is 3 elements and our subspace has 2 dimensions because of the 2 polynomials that make its basis?
the subspace has dimension 2 because it has a basis with 2 elements.
dimension = # of elements in a basis. that is the DEFINITION of dimension.
{x-2, 2x-4} is NOT a basis. why? because 2x-4 and x-2 are NOT linearly independent. 2x-4 = 2(x-2), so we can find a, and b not both 0, with
a(2x-4) + b(x-2) = 0. simply choose a = 1, and b = -2.
our subspace has dimension 2, because it has two polynomials that form a basis {x-2,(x-2)(x+3)} IS a basis. we've already seen it is linearly independent, let's see that it spans our subspace.
suppose f(2) = 0. well, we know x-2 is a factor. so f(x) = (x-2)g(x). since the degree of f is 2 or less, the degree of g must be 1 or less. so:
f(x) = (x-2)(cx+d) (we're going to asume we KNOW what c and d are). now, let's look at linear combinations of x-2, and (x-2)(x+3).
a(x-2) + b(x-2)(x+3) = (x-2)(bx + 3b + a). we want to choose a and b so that this comes out to be f. so let's pick:
a = d - 3c, b = c. now, let's see what a(x-2) + b(x-2)(x+3) works out to:
(d-3c)(x-2) + c(x-2)(x+3) = (x-2)(d-3c+cx+3c) = (x-2)(cx+d) = f(x). so it works, we can express f as a linear combination of x-2 and (x-2)(x+3).
so {x-2,(x-2)(x+3)} is one possible basis. there are LOTS of possible bases, they aren't unique. what is unique is the size of a basis, the dimension.
we have a basis of size 2, so every basis (of our subspace) has size 2. so if we find 2 linearly independent elements of our subspace, it's a basis.
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