Let $S$ a subspace and $V$ a vector space. Show that the additive identity of $S$ is the additive identity of $V$.

Question

Working on the book: Robert Messer. "Linear algebra - The gateway to mathematics" (p. 55)

16. Suppose $S$ is the subspace of a vector space $V$.

a. Show that the additive identity of $S$ is the additive identity of $V$.

This is my attempt to prove it:

• $0$ is the additive identity of $V$ and $0'$ is the additive identity of $S$.
  • Assume $v \in S$
  • $v + 0' = v$ ($0'$ is the additive identity of S)
  • $v + 0 = v$ ($0$ is the additive identity of V)
  • $\forall x(x \in V \to \exists! y(x+y=x))$ (0 is the unique additive identity of $V$)
  • $v \in V \to \exists! y(v+y=v)$
  • $v \in V$ (since S \subseteq V)
  • $\exists! y(v+y=v)$
    • $v+z=v \land \forall z'(z' \in V \land v+z'=z \to z'=z)$
    • $\forall z'(z' \in V \land v+z'=z \to z'=z)$
    • $0 \in V \land v+0=v \to 0=z$
    • $0' \in V \land v+0'=v \to 0'=z$
    • $0=0'$ (by transitivity)
    • $\vdots$

Is my proof skeleton correct ?

Would it suffice to show

• If $0$ is the additive identity of $V$ then $0'$ is the additive identity of $S$ and,
• If $0'$ is the additive identity of $V$ then $0$ is the additive identity of $S$ ?

Would appreciate some insight from a logic point of view.

Answer 1

My spiel is not too heavy on the logic, but I think it conveys the essential ideas:

Consider:

$\forall v \in V, \; 0_V + v = v + 0_V = v; \tag 1$

now let

$s \in S \subset V; \tag 2$

then

$s \in V, \tag 3$

whence, via (1),

$0_V + s = s; \tag 4$

also, by (2),

$0_S + s = s; \tag 5$

combining (4) and (5) we find

$0_V + s = 0_S + s; \tag 6$

we may now write

$(0_V + s) + (-s) = (0_S + s) + (-s), \tag 7$

from which

$0_V + (s + (-s)) = 0_S + (s + (-s)); \tag 8$

now,

$s + (-s) = 0_V, \tag 9$

so (8) becomes

$0_V + 0_V = 0_S + 0_V, \tag{10}$

and thus by virtue of (1),

$0_V = 0_V + 0_V = 0_S + 0_V = 0_S. \tag{11}$