A set of vectors \( {v_1, v_2, …, v_n} \)
in a vector space is said to be **linearly independent** if no vector in the set can be written as a linear combination of the others. Think of each vector as a direction in space. If a set of vectors is linearly independent, it means that none of these directions can be reached by walking along the others. Each vector adds a new dimension or direction that is not covered by any combination of the other vectors.

## Claim

*Claim*: If a set of vectors \( {v_1, v_2, v_3} \)
is linearly independent, then the set \( {v_1 - 4v_2, v_2, v_3} \)
is also linearly independent.

The claim provided above is an application or consequence of the original definition of linear independence. It illustrates a scenario where modifying one vector in a linearly independent set (in this case, subtracting \( 4v_2 \) from \( v_1 \)) does not introduce linear dependence among the vectors. This is because the modified vector \( (v_1 - 4v_2) \) still cannot be represented as a linear combination of the other vectors \( (v_2) \) and \( (v_3) \) in the new set. Let’s dive into the proof.

## Proof

**Step 1: Assume the Hypothesis**

Suppose that a set of vectors \( {v_1, v_2, v_3} \) is linearly independent.

**Step 2: Deduce the Conclusion**

The collection \( {v_1, v_2, v_3} \) being linearly independent means that the only linear relation on the collection is the trivial one. That is, if we have scalars \( \alpha_1, \alpha_2, \) and \( \alpha_3 \) such that:

\[ \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 = 0 \]

then it must follow that \( \alpha_1 = \alpha_2 = \alpha_3 = 0 \), indicating any such linear relation must be trivial.

**Step 3: Apply the Transformation**

Now, consider the set \( {v_1 - 4v_2, v_2, v_3} \). We want to show that this set is also linearly independent. Suppose there exist scalars \( \beta_1, \beta_2, \) and \( \beta_3 \) such that:

\[ \beta_1 (v_1 - 4v_2) + \beta_2 v_2 + \beta_3 v_3 = 0 \]

Expanding the above equation, we get:

\[ \beta_1 v_1 - 4\beta_1 v_2 + \beta_2 v_2 + \beta_3 v_3 = 0 \]

Rearranging, we find:

\[ \beta_1 v_1 + (-4\beta_1 + \beta_2) v_2 + \beta_3 v_3 = 0 \]

Since \( {v_1, v_2, v_3} \) is linearly independent, the only solution for this equation is when all the coefficients are zero:

\[ \beta_1 = 0, -4\beta_1 + \beta_2 = 0, \beta_3 = 0 \]

This implies that \( \beta_1 = \beta_2 = \beta_3 = 0 \). Therefore, the set \( {v_1 - 4v_2, v_2, v_3} \) is linearly independent.

## Conclusion

This proof demonstrates that linear independence is preserved even when one vector in a linearly independent set is modified by subtracting a scalar multiple of another vector from the same set.