Let S = v1...... v2 spans a vector space v some m-fold cartesian product of R. Prove in span a vector space V, some mfold Create a If xv = 0 for all v V, then x =0.

To prove the statement, let's break it down and carefully construct the argument.

### Problem Restatement
Let \( S = \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) be a set of vectors that spans a vector space \( V \), where \( V \) is some \( m \)-fold Cartesian product of \( \mathbb{R} \), i.e., \( V = \mathbb{R}^m \). We need to prove the following statement:

**If \( \mathbf{x} \in \mathbb{R}^m \) satisfies \( \mathbf{x} \cdot \mathbf{v} = 0 \) for all \( \mathbf{v} \in V \), then \( \mathbf{x} = \mathbf{0} \).**

### Proof

1. **Definition of the Span**:  
   Since \( S \) spans the vector space \( V \), any vector \( \mathbf{v} \in V \) can be written as a linear combination of the vectors in \( S \):
   \[
   \mathbf{v} = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \dots + a_n \mathbf{v}_n,
   \]
   where \( a_1, a_2, \dots, a_n \in \mathbb{R} \).

2. **Given Condition**:  
   We are given that \( \mathbf{x} \cdot \mathbf{v} = 0 \) for all \( \mathbf{v} \in V \). Substituting the expression for \( \mathbf{v} \) from the previous step, we get:
   \[
   \mathbf{x} \cdot (a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \dots + a_n \mathbf{v}_n) = 0.
   \]
   Using the distributive property of the dot product:
   \[
   a_1 (\mathbf{x} \cdot \mathbf{v}_1) + a_2 (\mathbf{x} \cdot \mathbf{v}_2) + \dots + a_n (\mathbf{x} \cdot \mathbf{v}_n) = 0.
   \]

3. **Coefficients Are Arbitrary**:  
   Since \( a_1, a_2, \dots, a_n \) are arbitrary real numbers and the above equation holds for all \( \mathbf{v} \in V \), the only way this can be true is if each individual dot product is zero:
   \[
   \mathbf{x} \cdot \mathbf{v}_i = 0 \quad \text{for all } i = 1, 2, \dots, n.
   \]

4. **Linear Independence of \( S \)**:  
   If the vectors \( \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \) are linearly independent, then the only solution to the above system of equations is \( \mathbf{x} = \mathbf{0} \). This is because if \( \mathbf{x} \) were non-zero, there would be some vector \( \mathbf{v}_i \) such that \( \mathbf{x} \cdot \mathbf{v}_i \neq 0 \), which contradicts the given condition.

   Even if the vectors in \( S \) are not linearly independent, the span condition still implies that \( \mathbf{x} \) must be orthogonal to every vector in the span, and hence to every vector in \( V \). This orthogonality to the entire vector space \( V \) implies that \( \mathbf{x} \) must be the zero vector, since the only vector orthogonal to every vector in \( \mathbb{R}^m \) is the zero vector.

### Conclusion
We have shown that if \( \mathbf{x} \) satisfies \( \mathbf{x} \cdot \mathbf{v} = 0 \) for all \( \mathbf{v} \in V \), then \( \mathbf{x} = \mathbf{0} \). This completes the proof.

---

Would you like any details or have any questions?

Here are some related questions you might consider:
1. What is the significance of linear independence in vector spaces?
2. Can a set of linearly dependent vectors span a vector space?
3. How does the dot product relate to orthogonality in vector spaces?
4. What are some other ways to prove that the zero vector is the only solution in such problems?
5. How do these concepts extend to complex vector spaces?

**Tip:** Always check if vectors in a set are linearly independent when determining if they span a vector space—this can significantly simplify proofs and calculations.

Learn how to prove that a set of vectors spans a vector space and how orthogonality leads to the zero vector in the context of vector spaces and Cartesian products of real numbers. Detailed step-by-step solution and theoretical background provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation