Math Problem Statement

solve this and write the whole question first then solve

Let V = P_{2024}(R), the vector space of all polynomials with real coefficients of degree at most 2024. Let W and U be two subspaces of V described by the following: W = {p(x) ∈ V : p(0) = p(1) = 0}, U = {p(x) ∈ V : p(0) = p(-1) = 0}. Find the dimensions of the subspaces W ∩ U and W + U.

Solution

The given problem is:


Let V=P2024(R)V = P_{2024}(\mathbb{R}), the vector space of all polynomials with real coefficients of degree at most 2024. Let WW and UU be two subspaces of VV described by the following:

W={p(x)V:p(0)=p(1)=0},W = \{p(x) \in V : p(0) = p(1) = 0\},

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

Find the dimensions of the subspaces WUW \cap U and W+UW + U.


Solution

Step 1: Understanding VV

The vector space V=P2024(R)V = P_{2024}(\mathbb{R}) consists of all polynomials of degree at most 2024, so: dim(V)=2025.\dim(V) = 2025.

Step 2: Subspace WW

In WW, the polynomials satisfy the conditions: p(0)=0andp(1)=0.p(0) = 0 \quad \text{and} \quad p(1) = 0.

These conditions imply two linear constraints on the coefficients of a general polynomial. Thus, the dimension of WW is: dim(W)=dim(V)2=20252=2023.\dim(W) = \dim(V) - 2 = 2025 - 2 = 2023.

Step 3: Subspace UU

In UU, the polynomials satisfy the conditions: p(0)=0andp(1)=0.p(0) = 0 \quad \text{and} \quad p(-1) = 0.

Similarly, these two conditions impose two linear constraints. Hence: dim(U)=dim(V)2=20252=2023.\dim(U) = \dim(V) - 2 = 2025 - 2 = 2023.

Step 4: Intersection WUW \cap U

The subspace WUW \cap U consists of all polynomials satisfying the conditions: p(0)=0,p(1)=0,andp(1)=0.p(0) = 0, \quad p(1) = 0, \quad \text{and} \quad p(-1) = 0.

These three conditions are linearly independent and impose three constraints. Therefore: dim(WU)=dim(V)3=20253=2022.\dim(W \cap U) = \dim(V) - 3 = 2025 - 3 = 2022.

Step 5: Sum of Subspaces W+UW + U

By the dimension formula for subspaces: dim(W+U)=dim(W)+dim(U)dim(WU).\dim(W + U) = \dim(W) + \dim(U) - \dim(W \cap U).

Substituting the values: dim(W+U)=2023+20232022=2024.\dim(W + U) = 2023 + 2023 - 2022 = 2024.


Final Answer

  1. The dimension of WUW \cap U is: dim(WU)=2022.\dim(W \cap U) = 2022.

  2. The dimension of W+UW + U is: dim(W+U)=2024.\dim(W + U) = 2024.


Let me know if you'd like any further explanation or details. Here are five related questions to deepen understanding:

  1. How does the dimension formula for subspaces help in solving similar problems?
  2. What are other possible interpretations of WUW \cap U and W+UW + U?
  3. How would the solution change if the degree of the polynomial were nn instead of 2024?
  4. Can we generalize these results to non-polynomial vector spaces?
  5. What if the conditions p(0),p(1),p(0), p(1), and p(1)p(-1) were dependent?

Tip: The dimension formula is a powerful tool for understanding relationships between subspaces. Use it systematically when working with sums and intersections!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Polynomials
Subspaces

Formulas

Dimension of a vector space: dim(V)
Dimension formula for subspaces: dim(W + U) = dim(W) + dim(U) - dim(W ∩ U)

Theorems

Dimension Theorem for Subspaces

Suitable Grade Level

Undergraduate Mathematics