Suppose z = u × v where × is the cross-product. Prove or provide
a counter example of the fact that u, v and z are linearly independent.

Let us first recall some facts about the cross product and linear independence.

### Definitions:
1. **Cross product**: Given two vectors \( u \) and \( v \) in \( \mathbb{R}^3 \), their cross product \( u \times v = z \) is a vector perpendicular to both \( u \) and \( v \).
   
2. **Linear independence**: A set of vectors \( \{u, v, z\} \) is linearly independent if the only solution to the equation
   \[
   c_1 u + c_2 v + c_3 z = 0
   \]
   is \( c_1 = c_2 = c_3 = 0 \).

### Approach to the problem:

We need to determine whether \( u \), \( v \), and \( z = u \times v \) are always linearly independent, or if there is a counterexample where they are linearly dependent.

#### Step 1: Special Case Analysis
Consider two simple cases that could break linear independence:

- **Case 1: \( u \) and \( v \) are parallel**. If \( u \) and \( v \) are parallel vectors, then their cross product is zero, i.e., \( u \times v = 0 \). Hence, the set \( \{u, v, z\} = \{u, v, 0\} \). 

  In this case, \( u \) and \( v \) are linearly dependent because \( v = k u \) for some scalar \( k \). Therefore, the set \( \{u, v, z\} \) is linearly dependent.

- **Case 2: \( u \) and \( v \) are non-parallel**. If \( u \) and \( v \) are not parallel, then \( u \times v \neq 0 \), and the resulting vector \( z = u \times v \) is perpendicular to both \( u \) and \( v \). Since \( z \) is perpendicular to the plane containing \( u \) and \( v \), the vectors \( u \), \( v \), and \( z \) should be linearly independent. 

   To prove this, assume there are scalars \( c_1, c_2, c_3 \) such that:
   \[
   c_1 u + c_2 v + c_3 (u \times v) = 0.
   \]
   Dot this equation with \( u \), and using the fact that \( u \cdot (u \times v) = 0 \), we get:
   \[
   c_1 (u \cdot u) + c_2 (u \cdot v) = 0.
   \]
   Similarly, dotting the equation with \( v \) gives:
   \[
   c_1 (v \cdot u) + c_2 (v \cdot v) = 0.
   \]
   Solving these two equations will lead to \( c_1 = 0 \) and \( c_2 = 0 \) if \( u \) and \( v \) are not parallel. Finally, since \( z = u \times v \), this also implies \( c_3 = 0 \), proving that the vectors are linearly independent.

### Conclusion:
- **If \( u \) and \( v \) are parallel**, \( u \), \( v \), and \( z = u \times v \) are linearly dependent because \( z = 0 \).
- **If \( u \) and \( v \) are not parallel**, \( u \), \( v \), and \( z \) are linearly independent.

Thus, the set \( \{u, v, z\} \) is **not always linearly independent**, and a counterexample occurs when \( u \) and \( v \) are parallel.

Would you like further details or explanations on any part of this?

### Related Questions:
1. What are some geometric interpretations of the cross product?
2. How can you use the determinant to check for linear independence?
3. How does the dot product relate to the concept of linear dependence?
4. In what cases does the magnitude of the cross product equal zero?
5. How would linear independence change in higher dimensions?

### Tip:
If two vectors are parallel, their cross product will always be zero, indicating that they are linearly dependent.

Suppose z = u × v where × is the cross-product. Prove or provide a counter example of the fact that u, v and z are linearly independent.

Explore how to prove or disprove the linear independence of vectors u, v, and their cross product z (u × v). This solution covers key concepts in linear algebra, including vector cross product properties and the conditions under which vectors are linearly dependent or independent. Suitable for undergraduate students.

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