Math Problem Statement

solve this.

Solve the problem of finding the dimensions of the subspaces W ∩ U and W + U given the vector space of polynomials V = P_{2024}(R), W = {p(x) ∈ V : p(0) = 0, p(1) = 0}, and U = {p(x) ∈ V : p(0) = 0, p(-1) = 0}.

Solution

The problem defines V=P2024(R)V = P_{2024}(\mathbb{R}), the vector space of all polynomials of degree at most 2024 with real coefficients. It further defines two subspaces WW and UU as follows:

  • W={p(x)V:p(0)=0,p(1)=0}W = \{ p(x) \in V : p(0) = 0, p(1) = 0 \},
  • U={p(x)V:p(0)=0,p(1)=0}U = \{ p(x) \in V : p(0) = 0, p(-1) = 0 \}.

We are tasked to find the dimensions of the subspaces WUW \cap U and W+UW + U.

Step 1: Dimension of VV

The dimension of VV, the vector space of all polynomials of degree at most 2024, is: dim(V)=2025.\dim(V) = 2025.

Step 2: Understanding WW

The subspace WW consists of polynomials in VV that satisfy: p(0)=0andp(1)=0.p(0) = 0 \quad \text{and} \quad p(1) = 0. This means p(x)p(x) must have roots at x=0x = 0 and x=1x = 1, so p(x)p(x) can be written as: p(x)=x(x1)q(x),p(x) = x(x-1)q(x), where q(x)q(x) is any polynomial of degree at most 20222022. Therefore: dim(W)=2023.\dim(W) = 2023.

Step 3: Understanding UU

The subspace UU consists of polynomials in VV that satisfy: p(0)=0andp(1)=0.p(0) = 0 \quad \text{and} \quad p(-1) = 0. This means p(x)p(x) must have roots at x=0x = 0 and x=1x = -1, so p(x)p(x) can be written as: p(x)=x(x+1)q(x),p(x) = x(x+1)q(x), where q(x)q(x) is any polynomial of degree at most 20222022. Therefore: dim(U)=2023.\dim(U) = 2023.

Step 4: Understanding WUW \cap U

The subspace WUW \cap U consists of polynomials in VV that satisfy: p(0)=0,p(1)=0,andp(1)=0.p(0) = 0, \quad p(1) = 0, \quad \text{and} \quad p(-1) = 0. This means p(x)p(x) must have roots at x=0x = 0, x=1x = 1, and x=1x = -1, so p(x)p(x) can be written as: p(x)=x(x1)(x+1)q(x),p(x) = x(x-1)(x+1)q(x), where q(x)q(x) is any polynomial of degree at most 20212021. Therefore: dim(WU)=2022.\dim(W \cap U) = 2022.

Step 5: Understanding W+UW + U

Using the dimension formula for the sum of two subspaces: dim(W+U)=dim(W)+dim(U)dim(WU).\dim(W + U) = \dim(W) + \dim(U) - \dim(W \cap U). Substituting the known dimensions: dim(W+U)=2023+20232022=2024.\dim(W + U) = 2023 + 2023 - 2022 = 2024.

Final Answers

  1. dim(WU)=2022\dim(W \cap U) = 2022,
  2. dim(W+U)=2024\dim(W + U) = 2024.

Would you like a detailed explanation of any specific step? Let me know!

Relative Questions:

  1. What happens if we consider V=Pn(R)V = P_n(\mathbb{R}) for a general nn instead of 2024?
  2. How does the dimension formula for W+UW + U apply to other vector spaces?
  3. Why does p(0)=0p(0) = 0 imply that p(x)p(x) must be divisible by xx?
  4. Can we use similar methods to compute dimensions in finite fields instead of real numbers?
  5. How do properties of polynomial roots relate to vector space subspaces?

Tip:

Always use the dimension formula for the sum of subspaces to verify results in vector space problems!

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Math Problem Analysis

Mathematical Concepts

Vector Spaces
Subspaces
Polynomials
Linear Algebra

Formulas

Dimension formula for subspace sum: dim(W + U) = dim(W) + dim(U) - dim(W ∩ U)

Theorems

Dimension Theorem for Subspaces

Suitable Grade Level

Undergraduate Mathematics (Linear Algebra)