From the axioms of a linear space, prove that 0v = 0 for each vector v

Sure, here is the proof:

1. We start with the axioms of a vector space. One of these axioms is that for every vector v in the vector space, there exists a scalar 0 such that 0v = 0.

2. To prove this, we can use another axiom of vector spaces: for every vector v, there exists an additive inverse -v such that v + (-v) = 0.

3. Now, consider the scalar multiplication 0v. We can write this as (1 - 1)v, since 0 = 1 - 1.

4. By the distributive property of scalar multiplication over vector addition, another axiom of vector spaces, we can rewrite this as 1v - 1v.

5. 1v is just v, by the definition of scalar multiplication. So we have v - v.

6. But from step 2, we know that v - v = v + (-v) = 0.

7. Therefore, we have shown that 0v = 0 for every vector v in the vector space.

This completes the proof.