A set of vectors $v_1,v_2,\dots ,v_n$ 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.

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