Math Problem Statement

Find a basis p(x), q(x) for the kernel of the linear transformation L : P3(x) -> R defined by L(f(x)) = f'(-6) - f(1) where P3(x) is the vector space of polynomials in x with degree less than 3

Solution

To find a basis for the kernel of the linear transformation L:P3(x)RL : P_3(x) \to \mathbb{R}, where L(f(x))=f(6)f(1)L(f(x)) = f'(-6) - f(1), follow these steps:

Step 1: Define the kernel of LL

The kernel of LL is the set of all polynomials f(x)P3(x)f(x) \in P_3(x) such that L(f(x))=0L(f(x)) = 0. That is: L(f(x))=f(6)f(1)=0.L(f(x)) = f'(-6) - f(1) = 0.

Step 2: Represent f(x)f(x) in P3(x)P_3(x)

A general polynomial in P3(x)P_3(x) has the form: f(x)=a0+a1x+a2x2,f(x) = a_0 + a_1x + a_2x^2, where a0,a1,a2Ra_0, a_1, a_2 \in \mathbb{R}.

Step 3: Compute f(6)f'(-6) and f(1)f(1)

The derivative of f(x)f(x) is: f(x)=a1+2a2x.f'(x) = a_1 + 2a_2x. Thus: f(6)=a1+2a2(6)=a112a2.f'(-6) = a_1 + 2a_2(-6) = a_1 - 12a_2. And: f(1)=a0+a1(1)+a2(12)=a0+a1+a2.f(1) = a_0 + a_1(1) + a_2(1^2) = a_0 + a_1 + a_2.

Step 4: Solve f(6)f(1)=0f'(-6) - f(1) = 0

Substituting into L(f(x))=0L(f(x)) = 0, we have: (a112a2)(a0+a1+a2)=0.(a_1 - 12a_2) - (a_0 + a_1 + a_2) = 0. Simplify: a112a2a0a1a2=0,a_1 - 12a_2 - a_0 - a_1 - a_2 = 0, a013a2=0.-a_0 - 13a_2 = 0. This simplifies further to: a0=13a2.a_0 = -13a_2.

Step 5: Write the kernel in parametric form

Let a2=ta_2 = t, where tRt \in \mathbb{R}. Then: a_0 = -13t, \quad a_1 = s \quad \text{(let a_1 = s, free variable)}. Thus: f(x)=a0+a1x+a2x2=13t+sx+tx2.f(x) = a_0 + a_1x + a_2x^2 = -13t + sx + tx^2. Factor tt and ss to express f(x)f(x) in terms of basis elements: f(x)=t(13+x2)+s(x).f(x) = t(-13 + x^2) + s(x).

Step 6: Extract the basis

The kernel is spanned by: p(x)=13+x2,q(x)=x.p(x) = -13 + x^2, \quad q(x) = x.

Final Answer:

The basis for the kernel is: {p(x),q(x)}={13+x2,x}.\{p(x), q(x)\} = \{-13 + x^2, x\}.

Would you like a deeper explanation or additional examples?

  1. What is the geometric interpretation of the kernel of a linear transformation?
  2. How does the kernel relate to the rank-nullity theorem?
  3. Could this kernel basis change if the degree of P3(x)P_3(x) were higher?
  4. How would the calculation change if L(f(x))=f(c)f(d)L(f(x)) = f'(c) - f(d) for general cc and dd?
  5. What is the relationship between the kernel of LL and the space P3(x)P_3(x)?

Tip: To verify your basis, plug it back into L(f(x))=0L(f(x)) = 0 and confirm that the transformation evaluates to zero for each basis element.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Kernels of Linear Transformations
Polynomial Functions

Formulas

f(x) = a_0 + a_1x + a_2x^2
f'(x) = a_1 + 2a_2x
f'(-6) = a_1 - 12a_2
f(1) = a_0 + a_1 + a_2
L(f(x)) = f'(-6) - f(1)

Theorems

Rank-Nullity Theorem

Suitable Grade Level

Undergraduate (Linear Algebra Course)