Proof of Linear Functionals Respect Differences, and Coefficients Along a Nonzero Vector Are Unique

Proof of (i). By the definition of the difference, $u-v=u+(-v)$, and $-v=(-1)v$ by Elementary Identities in a Vector Space (iv). Since $W$ is a linear subspace, $(-1)v\in W$ by Linear Subspace of a Real Vector Space (iii), and then $u-v=u+(-1)v\in W$ by Linear Subspace of a Real Vector Space (ii). By Linear Functional on a Linear Subspace (i) and Linear Functional on a Linear Subspace (ii),

$$g(u-v)=g\bigl(u+(-1)v\bigr)=g(u)+g\bigl((-1)v\bigr)=g(u)+(-1)\,g(v)=g(u)-g(v),$$

the last step being arithmetic in $\mathbb{R}$.

Proof of (ii). Using Vector Space over a Field (viii), the identity $(-t')x_{0}=-(t'x_{0})$ of Elementary Identities in a Vector Space (iv) and $w-w=0_{V}$ for $w\in V$ from Elementary Identities in a Vector Space (vii), we get

$$(t-t')x_{0}=\bigl(t+(-t')\bigr)x_{0}=tx_{0}+(-t')x_{0}=tx_{0}-t'x_{0}=t'x_{0}-t'x_{0}=0_{V},$$

where the fourth equality uses the hypothesis $tx_{0}=t'x_{0}$. By Elementary Identities in a Vector Space (xii), $t-t'=0$ or $x_{0}=0_{V}$; as $x_{0}\neq0_{V}$, $t-t'=0$, that is, $t=t'$.