Theorem on the Linear Independence of Vectors

If a set of vectors {v₁, v₂, ..., v_n} are linearly independent, then each vector is distinct from the zero vector $$ \vec{v}_1 \ne \vec{0} \\ \vec{v}_2 \ne \vec{0} \\ \vdots \\ \vec{v}_n \ne \vec{0} $$

The Proof

Consider n linearly independent vectors {v₁, v₂, ..., v_n}.

For the sake of contradiction, assume that one of the vectors, v_k, is a zero vector.

$$ \vec{v}_k = \vec{0} $$

Now, consider the linear combination of these vectors:

$$ \lambda_1 \vec{v}_1 + \lambda_1 \vec{v}_2 + ... + \lambda_k \vec{v}_k + ... + \lambda_n \vec{v}_n $$

Set all scalar coefficients λ₁, λ₂, ..., λ_n to zero, except for the scalar coefficient λ_k which is nonzero (λ_k≠0).

$$ \lambda_1 = \lambda_2 = ... = \lambda_n = 0 $$

$$ \lambda_k \ne 0 $$

Since v_k is a zero vector, the linear combination of the vectors equals the zero vector:

$$ 0 \cdot \vec{v}_1 + 0 \cdot \vec{v}_2 + ... + \lambda_k \vec{v}_k + ... + 0 \cdot \vec{v}_n = 0 $$

However, this contradicts our initial assumption, implying that the vectors are linearly dependent.

Contrary to our initial hypothesis, where the vectors are presumed linearly independent.

Note. The linear combination of linearly independent vectors equals zero only when all scalar coefficients are zero (the trivial solution). In this case, λ_k is nonzero, yet the linear combination is still zero. Thus, the vectors are not linearly independent.

This demonstrates that no linearly independent vector can be a zero vector.

And so on.