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.