### Linear Independence Proof - Theorem and Explanation

#### Theorem (Linear Independence Criterion):
A set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) in a vector space \( V \) over a field \( F \) is **linearly independent** if and only if the equation

\[
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_n \mathbf{v}_n = \mathbf{0}
\]

implies that all the scalars \( c_1, c_2, \dots, c_n \) are zero, i.e., \( c_1 = c_2 = \dots = c_n = 0 \).

#### Proof:

1. **Assume the vectors are linearly independent**:
   - By definition, the set \( \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) is linearly independent if the only solution to the equation 

   \[
   c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_n \mathbf{v}_n = \mathbf{0}
   \]

   is \( c_1 = c_2 = \dots = c_n = 0 \).
   - To prove this, consider the equation above. Suppose \( c_1, c_2, \dots, c_n \) are not all zero. This would imply that at least one of the coefficients \( c_i \) is non-zero. 
   - If at least one \( c_i \) is non-zero, the linear combination of \( \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \) cannot be zero (because the vectors are linearly independent). However, this contradicts our assumption that \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_n \mathbf{v}_n = \mathbf{0} \).
   - Therefore, the assumption that \( c_1, c_2, \dots, c_n \) are not all zero leads to a contradiction, implying that \( c_1 = c_2 = \dots = c_n = 0 \) is the only solution.

2. **Conversely, assume that the only solution to the equation is \( c_1 = c_2 = \dots = c_n = 0 \)**:
   - We need to show that the vectors are linearly independent.
   - Suppose the set \( \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) were linearly dependent. This would mean there exists a non-trivial linear combination of the vectors that equals zero, i.e., there are scalars \( c_1, c_2, \dots, c_n \), not all zero, such that 

   \[
   c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_n \mathbf{v}_n = \mathbf{0}.
   \]
   - But this contradicts the assumption that the only solution is \( c_1 = c_2 = \dots = c_n = 0 \).
   - Hence, the vectors must be linearly independent.

Therefore, a set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) is linearly independent if and only if the only solution to the equation 

\[
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_n \mathbf{v}_n = \mathbf{0}
\]

is \( c_1 = c_2 = \dots = c_n = 0 \).

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Would you like further details or have any questions?

Here are 5 related questions to explore this concept further:

1. How does the concept of linear independence differ from linear dependence?
2. What is the geometric interpretation of linear independence in \( \mathbb{R}^2 \) and \( \mathbb{R}^3 \)?
3. How can you use a matrix to test the linear independence of a set of vectors?
4. What is the significance of the rank of a matrix in determining linear independence?
5. Can you explain the relationship between linear independence and the span of a set of vectors?

**Tip:** When proving linear independence, remember that the key is showing that the only linear combination of the vectors that equals zero is the trivial combination where all coefficients are zero.

Understand the concept of linear independence in vector spaces with a detailed proof of the Linear Independence Criterion theorem. Explore how sets of vectors are determined to be linearly independent and the implications in linear algebra.

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