Let S = v1...... v2 spans a vector space v some m-fold cartesian product of R prove If xvk = 0 for all vk S, then x = 0.

Let's formalize and prove the statement:

### 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 want to prove the following statement:

**If \( \mathbf{x} \cdot \mathbf{v}_k = 0 \) for all \( \mathbf{v}_k \in S \), then \( \mathbf{x} = \mathbf{0} \).**

### Proof

1. **Given Condition**:
   We are given that \( \mathbf{x} \cdot \mathbf{v}_k = 0 \) for all \( k = 1, 2, \dots, n \). In other words, \( \mathbf{x} \) is orthogonal to each vector \( \mathbf{v}_k \) in the set \( S \).

2. **Spanning Condition**:
   The set \( S \) spans the vector space \( V \), meaning 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 \( a_1, a_2, \dots, a_n \in \mathbb{R} \).

3. **Orthogonality Implies Zero**:
   Consider any vector \( \mathbf{v} \in V \). Since \( \mathbf{x} \) is orthogonal to each \( \mathbf{v}_k \), the dot product of \( \mathbf{x} \) with \( \mathbf{v} \) is:
   \[
   \mathbf{x} \cdot \mathbf{v} = \mathbf{x} \cdot (a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \dots + a_n \mathbf{v}_n).
   \]
   By the distributive property of the dot product, this can be expanded as:
   \[
   \mathbf{x} \cdot \mathbf{v} = a_1 (\mathbf{x} \cdot \mathbf{v}_1) + a_2 (\mathbf{x} \cdot \mathbf{v}_2) + \dots + a_n (\mathbf{x} \cdot \mathbf{v}_n).
   \]
   But since \( \mathbf{x} \cdot \mathbf{v}_k = 0 \) for all \( k \), it follows that:
   \[
   \mathbf{x} \cdot \mathbf{v} = 0 \quad \text{for all } \mathbf{v} \in V.
   \]

4. **Implication for \( \mathbf{x} \)**:
   The only vector \( \mathbf{x} \) in \( \mathbb{R}^m \) that is orthogonal to every vector in \( V \) (which spans \( \mathbb{R}^m \)) is the zero vector. Therefore, \( \mathbf{x} \) must be \( \mathbf{0} \).

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

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Would you like any more details or have any questions?

Here are some related questions to consider:
1. How does the dimensionality of \( V = \mathbb{R}^m \) affect the spanning set \( S \)?
2. What is the role of the dot product in determining vector orthogonality?
3. Can this result be generalized to vector spaces over fields other than \( \mathbb{R} \)?
4. How would the proof change if \( S \) was not a spanning set?
5. What are some practical applications of this result in linear algebra?

**Tip:** In linear algebra, proving that a vector is the zero vector often hinges on showing it is orthogonal to a set of vectors that spans the entire space.

Explore how to prove that if the dot product of a vector with every vector in a spanning set of a vector space is zero, then the vector must be the zero vector. Detailed mathematical proof and implications discussed.

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