To prove the statement $(−c)v = −(cv)$ for a vector $v \in V$ in a vector space $V$ over a field $F$ and a scalar $c \in F$, we can proceed as follows:

Proof:

1. Recall the properties of scalar multiplication:

   • Distributive property: $a(v_1 + v_2) = av_1 + av_2$ for $a \in F$ and $v_1, v_2 \in V$.
   • Distributive property over field addition: $(a + b)v = av + bv$ for $a, b \in F$ and $v \in V$.
   • Scalar multiplication associativity: $(ab)v = a(bv)$ for $a, b \in F$ and $v \in V$.
   • Scalar multiplication identity: $1v = v$ for $v \in V$ and $1 \in F$ (where $1$ is the multiplicative identity in $F$).
   • Scalar multiplication with zero: $0v = 0_V$ (where $0_V$ is the zero vector in $V$).

2. Consider the expression $(−c)v$:

   We know that $−c$ is the additive inverse of $c$ in $F$. By definition of scalar multiplication and the properties of the field, we have: $(−c)v + (cv) = (−c + c)v = 0v = 0_V$ This equation tells us that $(−c)v$ is the additive inverse of $cv$. In other words: $(−c)v = −(cv)$ where $−(cv)$ denotes the additive inverse of the vector $cv$.

Conclusion:

Thus, we have shown that $(−c)v = −(cv)$ for any vector $v \in V$ and scalar $c \in F$.

Would you like more details or have any questions?

Here are 5 related questions you might consider:

1. How does the proof change if $V$ is over a ring instead of a field?
2. Can you show that $0 \cdot v = 0_V$ using the properties of a vector space?
3. What are the implications of the scalar multiplication properties on the structure of the vector space?
4. How does the proof apply if $v$ is the zero vector?
5. What role does the distributive property play in the proof?

Tip: When proving identities in vector spaces, always rely on the axioms and properties of vector spaces, such as distributive properties and associativity, to derive the required result.