Assignment 6, Solution to problem 1

Sorry these are going up a little late. Make sure you can understand, especially question 1.

Problem 1

(i)

P₄ is a vector space. Here is a check of the first four conditions given in lecture 10

• (1) If

  a₁x⁴+ b₁x³+ c₁x²+ d₁x+ e₁

  and a₂x⁴+ b₂x³+ c₂x²+ d₂x+ e₂

  are polynomials in P₄, then so is their sum:

  (a₁x⁴+ b₁x³+ c₁x²+ d₁x+ e₁) + (a₂x⁴+ b₂x³+ c₂x²+ d₂x+ e₂)=

  (a₁+a₂) x⁴+ (b₁+b₂) x³+ (c₁+c₂) x²+ (d₁+d₂) x+ (e₁+e₂)

• (2) If f= a₁x⁴+ b₁x³+ c₁x²+ d₁x+ e₁ is a polynomial in P₄, and k is any real number, then k*f is also in P₄:

  k(ax⁴+ bx³+ cx²+ dx+ e)= kax⁴+ kbx³+ kcx²+ kdx+ ke

• (3) The zero polynomial is in P₄. Just take a=b=c=d=e=0:

  0x⁴+ 0x³+ 0x²+ 0x+ 0e=0

• (4) If f= a₁x⁴+ b₁x³+ c₁x²+ d₁x+ e₁ is a polynomial in P₄, then -f is also in P₄:

  -(ax⁴+ bx³+ cx²+ dx+ e)= -ax⁴- bx³- cx²- dx-e

(ii) A basis for P₄:

We can take the following as a basis of P₄:

If we compare this with the vector space R⁵, we have a correspondance:

(iii) and (iv) Subspaces of P₄

(a) The set of polynomials with the coefficient of x² =7

This is not a subspace of P₄.

For example, the following polynomials are in this set, since they have coefficient of x² equal to 7:

2x⁴+ 3x³+ 7x²+ 2

x⁴+ 7x²+ 22x

but if you add them together, you get a polynomial which is not in the set of polynomials with the coefficient of x² =7:

(b) The set of polynomials f in P₄ with f(0)=3

This is also not a subspace of P₄.

For example, if f(0)=3, and g(0)=3, f and g are both in this set, but their sum, f+g, is not, since (f+g)(0)=f(0)+g(0)=3+3=6, so (f+g)(0) does not equal 3.

For a more explicit example, we could take f=x+3 and g=2x⁴+3, and then f+g=2x⁴+x+6, so f+g is not in the set of polynomials which have value 3 when x=0.

(c) The set of polynomials with P₄ f(1)=f(2), and f(3)=0

This is a subspace. Check:

If f and g are in this set, and k is any real number,

• (f+g)(1)=f(1)+g(1)=f(2)+g(2)=(f+g)(2)
• (f+g)(3)=f(3)+g(3)=0+0=0
• kf(1)=kf(2)
• kf(3)=k*0=0

So from these caluclations, f+g and kf are also in the set, so it is a subspace. (The other properties of subspaces will be very easy to check if you want to check them).

To find a basis of this set, we need to look at what conditions are imposed on the coeffieicnts.

If f=(a,b,c,d,e)_B, and f(1)=f(2), then

So

So

If f(3)=0, then

So to find the polynomials with f(1)=f(2) and f(3)=0 means to find the solutions of the linear system:

We get:

So (a,b,c,d,e)=(0.1111c+0.037d-0.0432e,-0.6667c-0.2222d-0.0926c,c,d,e) =c(0.1111,-0.6667,1,0,0) +d(0.037,-0.2222,0,1,0) +e(-0.0432,-0.0926,0,0,1)

So a basis is given by:

As polynomials, the basis is:

Please send comments, questions, suggestions, corrections, ideas, to me.