Let's analyze and solve the mathematical problem written in the uploaded image. The problem involves subspaces of $P_3(x)$, which is the space of polynomials of degree at most 3.

Problem Statement

Let $U$ and $V$ be subspaces of $P_3(x)$ defined as follows:

1. $U = \{ p(x) \in P_3(x) : p(1) = 0 \}$
2. $V = \{ p(x) \in P_3(x) : p(-1) = 0 \}$

Our goal is likely to:

• Analyze these subspaces in terms of their properties, dimensions, and possibly their intersection.

Step 1: Understand the Subspaces

Definition of $P_3(x)$:

The space $P_3(x)$ consists of all polynomials of degree at most 3: $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3, \quad a_0, a_1, a_2, a_3 \in \mathbb{R}.$ The dimension of $P_3(x)$ is 4, since it is spanned by the basis $\{1, x, x^2, x^3\}$.

Subspace $U$:

The condition $p(1) = 0$ implies: $p(1) = a_0 + a_1(1) + a_2(1^2) + a_3(1^3) = a_0 + a_1 + a_2 + a_3 = 0.$ This is a single linear constraint on the coefficients of $p(x)$. Since $U$ is a subspace of $P_3(x)$, its dimension is $\dim(P_3(x)) - 1 = 4 - 1 = 3$.

A basis for $U$ can be found by imposing the condition $a_0 + a_1 + a_2 + a_3 = 0$, but we’ll address this explicitly if needed.

Subspace $V$:

The condition $p(-1) = 0$ implies: $p(-1) = a_0 + a_1(-1) + a_2(-1)^2 + a_3(-1)^3 = a_0 - a_1 + a_2 - a_3 = 0.$ This is another single linear constraint on the coefficients of $p(x)$. Similarly, $V$ is a subspace of $P_3(x)$ with $\dim(V) = 3$.

Step 2: Intersection of $U$ and $V$

The intersection $U \cap V$ consists of all polynomials in $P_3(x)$ that satisfy both $p(1) = 0$ and $p(-1) = 0$. These two conditions give a system of linear equations:

1. $a_0 + a_1 + a_2 + a_3 = 0$ (from $p(1) = 0$),
2. $a_0 - a_1 + a_2 - a_3 = 0$ (from $p(-1) = 0$).

Let’s solve this system for $a_0, a_1, a_2, a_3$:

• Adding the two equations: $2a_0 + 2a_2 = 0 \implies a_0 + a_2 = 0 \implies a_2 = -a_0.$
• Subtracting the second equation from the first: $2a_1 + 2a_3 = 0 \implies a_1 + a_3 = 0 \implies a_3 = -a_1.$

Substituting $a_2 = -a_0$ and $a_3 = -a_1$, we can express $p(x)$ as: $p(x) = a_0(1 - x^2) + a_1(x - x^3).$

Thus, the intersection $U \cap V$ is spanned by the basis $\{1 - x^2, x - x^3\}$, and its dimension is 2.

Step 3: Dimension Formula Verification

The formula for the dimensions of subspaces is: $\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V).$

Since $\dim(U) = 3$, $\dim(V) = 3$, and $\dim(U \cap V) = 2$, we have: $\dim(U + V) = 3 + 3 - 2 = 4.$

This confirms that $U + V = P_3(x)$, as the dimension of $P_3(x)$ is 4.

Final Results

1. $\dim(U) = 3$,
2. $\dim(V) = 3$,
3. $\dim(U \cap V) = 2$,
4. $U + V = P_3(x)$.

Would you like a deeper explanation of any step or further expansion?

Related Questions

1. How do you find a basis for $U$ or $V$ explicitly?
2. Can you explain why $U + V = P_3(x)$?
3. How do we generalize these results to higher-degree polynomial spaces $P_n(x)$?
4. What is the geometric interpretation of subspaces $U$ and $V$ in $P_3(x)$?
5. How does the rank-nullity theorem apply in this context?

Tip

When working with subspaces, always verify the dimensions using the dimension formula to ensure consistency!