Math Problem Statement

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.

Solution

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

Definitions:

  1. Cross product: Given two vectors uu and vv in R3\mathbb{R}^3, their cross product u×v=zu \times v = z is a vector perpendicular to both uu and vv.

  2. Linear independence: A set of vectors {u,v,z}\{u, v, z\} is linearly independent if the only solution to the equation c1u+c2v+c3z=0c_1 u + c_2 v + c_3 z = 0 is c1=c2=c3=0c_1 = c_2 = c_3 = 0.

Approach to the problem:

We need to determine whether uu, vv, and z=u×vz = 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: uu and vv are parallel. If uu and vv are parallel vectors, then their cross product is zero, i.e., u×v=0u \times v = 0. Hence, the set {u,v,z}={u,v,0}\{u, v, z\} = \{u, v, 0\}.

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

  • Case 2: uu and vv are non-parallel. If uu and vv are not parallel, then u×v0u \times v \neq 0, and the resulting vector z=u×vz = u \times v is perpendicular to both uu and vv. Since zz is perpendicular to the plane containing uu and vv, the vectors uu, vv, and zz should be linearly independent.

    To prove this, assume there are scalars c1,c2,c3c_1, c_2, c_3 such that: c1u+c2v+c3(u×v)=0.c_1 u + c_2 v + c_3 (u \times v) = 0. Dot this equation with uu, and using the fact that u(u×v)=0u \cdot (u \times v) = 0, we get: c1(uu)+c2(uv)=0.c_1 (u \cdot u) + c_2 (u \cdot v) = 0. Similarly, dotting the equation with vv gives: c1(vu)+c2(vv)=0.c_1 (v \cdot u) + c_2 (v \cdot v) = 0. Solving these two equations will lead to c1=0c_1 = 0 and c2=0c_2 = 0 if uu and vv are not parallel. Finally, since z=u×vz = u \times v, this also implies c3=0c_3 = 0, proving that the vectors are linearly independent.

Conclusion:

  • If uu and vv are parallel, uu, vv, and z=u×vz = u \times v are linearly dependent because z=0z = 0.
  • If uu and vv are not parallel, uu, vv, and zz are linearly independent.

Thus, the set {u,v,z}\{u, v, z\} is not always linearly independent, and a counterexample occurs when uu and vv 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.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Calculus

Formulas

Cross product formula: u × v = z
Linear independence equation: c1u + c2v + c3z = 0

Theorems

Linear Independence Theorem
Cross Product Properties

Suitable Grade Level

College level (Undergraduate)