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We are given the subspaces \( U \) and \( V \) of \( \mathbb{R}^4 \) defined as:

\[
U = \text{span}\{ 3 - x + 4x^2 + x^3, 5 + 5x^2 + x^3, 5 - 5x + 10x^2 + 3x^3 \}
\]
\[
V = \text{span}\{ 9 - 3x + 3x^2 + 2x^3, 5 - x + 2x^2 + x^3, 6 + 4x^2 + x^3 \}
\]

We need to determine:
1. A basis for \( U \)
2. A basis for \( V \)
3. A basis for \( U + V \)
4. A basis for \( U \cap V \)
5. The dimensions of these subspaces.

### Step 1: Write the vectors in matrix form
We can write the given vectors in each subspace as coordinate vectors in \( \mathbb{R}^4 \). Each vector corresponds to a polynomial of degree at most 3, so we represent each vector as a 4-tuple \((a_0, a_1, a_2, a_3)\) corresponding to the polynomial \( a_0 + a_1x + a_2x^2 + a_3x^3 \).

For \( U \):
\[
3 - x + 4x^2 + x^3 \rightarrow (3, -1, 4, 1)
\]
\[
5 + 5x^2 + x^3 \rightarrow (5, 0, 5, 1)
\]
\[
5 - 5x + 10x^2 + 3x^3 \rightarrow (5, -5, 10, 3)
\]

For \( V \):
\[
9 - 3x + 3x^2 + 2x^3 \rightarrow (9, -3, 3, 2)
\]
\[
5 - x + 2x^2 + x^3 \rightarrow (5, -1, 2, 1)
\]
\[
6 + 4x^2 + x^3 \rightarrow (6, 0, 4, 1)
\]

### Step 2: Find the basis for \( U \)
We will now perform row reduction to find the basis for \( U \) by checking the linear independence of the vectors:
\[
\text{Matrix for } U = \begin{pmatrix}
3 & -1 & 4 & 1 \\
5 & 0 & 5 & 1 \\
5 & -5 & 10 & 3
\end{pmatrix}
\]

### Step 3: Find the basis for \( V \)
Similarly, we need to check the linear independence of the vectors for \( V \):
\[
\text{Matrix for } V = \begin{pmatrix}
9 & -3 & 3 & 2 \\
5 & -1 & 2 & 1 \\
6 & 0 & 4 & 1
\end{pmatrix}
\]

After that, we can combine the bases of \( U \) and \( V \) to find \( U + V \), and then find their intersection \( U \cap V \).

I'll proceed with row reduction on these matrices.Here are the results from the row reduction:

1. **Basis and Dimension of \( U \):**
   - The rank of the matrix for \( U \) is 2, meaning the dimension of \( U \) is 2. This implies that the vectors span a 2-dimensional subspace.
   - A basis for \( U \) can be derived from the two linearly independent rows after row reduction.

2. **Basis and Dimension of \( V \):**
   - The rank of the matrix for \( V \) is 3, meaning the dimension of \( V \) is 3. So, the vectors in \( V \) span a 3-dimensional subspace.
   - A basis for \( V \) consists of the three independent rows.

3. **Basis and Dimension of \( U + V \):**
   - The rank of the combined matrix \( U \cup V \) is 4, meaning the dimension of \( U + V \) is 4. This indicates that the union of \( U \) and \( V \) spans the entire space \( \mathbb{R}^4 \).
   
4. **Dimension of \( U \cap V \):**
   - Using the formula for the intersection of subspaces:
     \[
     \dim(U \cap V) = \dim(U) + \dim(V) - \dim(U + V)
     \]
     \[
     \dim(U \cap V) = 2 + 3 - 4 = 1
     \]
   So, the dimension of \( U \cap V \) is 1, indicating that they share a 1-dimensional subspace.

If you'd like, I can further detail the specific basis vectors for each subspace. Let me know if you want to explore any part in more detail!

---

**Relative Questions:**
1. How do you find the basis for a subspace from row-reduced matrices?
2. What is the significance of the dimension of the intersection of subspaces?
3. Can a subspace of \( \mathbb{R}^4 \) have more than 4 dimensions?
4. How do you determine the linear independence of vectors?
5. Why is the dimension of \( U + V \) not simply the sum of the dimensions of \( U \) and \( V \)?

**Tip:** When finding the intersection \( U \cap V \), always check if the vectors in \( U \) can be expressed as linear combinations of vectors in \( V \) (and vice versa).

Let the subspaces U and V of R^4 defined as U = span{3 − x + 4x^2 + x^3, 5 + 5x^2 + x^3, 5 − 5x + 10x^2 + 3x^3} and V = span{9 − 3x + 3x^2 + 2x^3, 5 − x + 2x^2 + x^3, 6 + 4x^2 + x^3}. Then determine a basis and dimension of U, V, U + V, and U ∩ V.

Learn how to find the basis and dimension of subspaces U and V in R^4. This problem covers linear algebra concepts such as vector spaces, span, and intersection. Suitable for college-level students studying advanced linear algebra.

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