We are tasked with proving the equivalence:

$\forall x \in V, \forall \alpha \in K, \quad \alpha \cdot x = 0_V \iff \alpha = 0_K \text{ or } x = 0_V$

Here, $V$ is a vector space over the field $K$, $x \in V$ is a vector, and $\alpha \in K$ is a scalar. The notation $0_V$ represents the zero vector in $V$, and $0_K$ represents the zero scalar in $K$.

Proof

$\Rightarrow$ Direction (If $\alpha \cdot x = 0_V$, then $\alpha = 0_K$ or $x = 0_V$)

1. Assume $\alpha \cdot x = 0_V$.
2. We want to show that this implies $\alpha = 0_K$ or $x = 0_V$.

There are two possibilities:

• Case 1: $\alpha = 0_K$

  If $\alpha = 0_K$, then clearly $\alpha \cdot x = 0_K \cdot x = 0_V$ for any $x \in V$, so the statement holds.

• Case 2: $\alpha \neq 0_K$

  Now suppose $\alpha \neq 0_K$. We need to show that in this case, $x = 0_V$. Since $V$ is a vector space, it is closed under scalar multiplication. Consider the scalar multiplication $\alpha^{-1} \cdot (\alpha \cdot x)$. By the properties of scalar multiplication:

  $\alpha^{-1} \cdot (\alpha \cdot x) = (\alpha^{-1} \alpha) \cdot x = 1_K \cdot x = x$

  But also, since $\alpha \cdot x = 0_V$, we have:

  $\alpha^{-1} \cdot 0_V = 0_V$

  Thus, we find $x = 0_V$. Therefore, $\alpha \cdot x = 0_V$ implies $x = 0_V$ when $\alpha \neq 0_K$.

Thus, we conclude that if $\alpha \cdot x = 0_V$, then either $\alpha = 0_K$ or $x = 0_V$.

$\Leftarrow$ Direction (If $\alpha = 0_K$ or $x = 0_V$, then $\alpha \cdot x = 0_V$)

1. Case 1: $\alpha = 0_K$

   If $\alpha = 0_K$, then for any $x \in V$, we have:

   $\alpha \cdot x = 0_K \cdot x = 0_V$

   So in this case, $\alpha \cdot x = 0_V$.

2. Case 2: $x = 0_V$

   If $x = 0_V$, then for any $\alpha \in K$, we have:

   $\alpha \cdot 0_V = 0_V$

   So in this case as well, $\alpha \cdot x = 0_V$.

Thus, we conclude that if $\alpha = 0_K$ or $x = 0_V$, then $\alpha \cdot x = 0_V$.

Conclusion

We have proven both directions:

$\forall x \in V, \forall \alpha \in K, \quad \alpha \cdot x = 0_V \iff \alpha = 0_K \text{ or } x = 0_V$

This completes the proof.

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Do you need further details or have any questions? Here are a few related questions:

1. How does the axiom of scalar multiplication in a vector space ensure that $\alpha \cdot x = 0_V$ implies either $\alpha = 0_K$ or $x = 0_V$?
2. Can this result be extended to other types of algebraic structures like modules or algebras?
3. How does the property $\alpha \cdot x = 0_V$ relate to the linear independence of vectors?
4. What are some real-world examples where this result is applied, such as in physics or computer science?
5. How does this theorem interact with the concept of null spaces in linear algebra?

Tip: It's important to remember that this result highlights the uniqueness of the zero vector and the zero scalar in the context of a vector space.