Thread: span Polynomial-concept and method confusion

Show that X spans P2
P_{2 consists of all polynomials of degree less than or equal to 2. let X={p1, p2,p3}} where p1=x+2
p2=3-x
p3=x²–x
It seems that I am still confused with the concept of Span. For span it is not necessary that it should be linearly independent. It could even be dependent.Span just means any linear combination?Are my thoughts correct or ..?
So What is the method for this question?
But then this question is for up to p2 excluding p3 what should i do?thanks.

Hey n22.

The concept of span is to find a set of vectors where you take an input set and produce an output set of vectors whereby you can write all the input vectors as linear combinations of your output vectors.

So since you have P2 (all polynomials of degree 2) you need to show that given the set X you can write all polynomials P2 using linear combinations of the vectors in X.

Hint: Show that all vectors in X are independent and show that as a result all vectors in P2 can be written as a linear combination.

Originally Posted by chiro

Hey n22.

The concept of span is to find a set of vectors where you take an input set and produce an output set of vectors whereby you can write all the input vectors as linear combinations of your output vectors.

So since you have P2 (all polynomials of degree 2) you need to show that given the set X you can write all polynomials P2 using linear combinations of the vectors in X.

Hint: Show that all vectors in X are independent and show that as a result all vectors in P2 can be written as a linear combination.

What chiro said, or show that each of the functions in a known basis (I assume you do know of a basis for P2) can be written as linear combinations of of your given functions.

.

In other words, any member of P2 can be written $\displaystyle ax^2+ bx+ c$. To show that $\displaystyle \{x+ 2, 3- x, x^2- x\}$ spans the space you must show that you can find number, p, q, and r such that $\displaystyle p(x+ 2)+ q(3- x)+ r(x^2- x)= ax^2+ bx+ c$ for any a, b, c.

The obvious thing to do is multiply out the left side and "combine like terms": $\displaystyle rx^2+ (p- q- r)x+ (2p+ 3q)= ax^2+ bx+ c$.
Using the fact that $\displaystyle \{x^2, x, 1\}$ is a basis for P2, we must have r= a, p- q- r= b, and 2p+ 3q= c. Since r= a, p- q- r= p- q- a= c so p- q= a+ c. Multiply that by 3 and add to 2p+ 3q= c: (3p- 3q)+ (2p+ 3q)= 3a+ 3c+c so 5p= 3a+ 4c. p= (3a+ 4c)/5. We can then work back to find b and c. Since this has an answer the set spans the space.

By the way, any basis for a vector space has three properties:
1) It spans the space.
2) It is independent.
3) The number of vectors in the basis is equal to the dimension of the space.
And any two of these implie the third!

Since P2 is obviously three dimensional and these three vectors span the space, they are also independent and form a basis for the space.

(I said above that the fact that there exist a solution to the system of equations implies that the vector span the space. The fact that there is a unique solution implies that they are also independent and this is a basis.)