Non standard vector addition

Question

If addition was defined as $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1, 0)$ over a a set $V$, the set of all ordered pairs of real numbers, does that special addition only apply to ordered pairs and vectors or would it also apply to scalars? If it would apply to scalars, please explain. I have trouble visualizing how that would apply to scalars. Thanks!

Answer 1

Given that you're writing vectors in $V$ as $(a_1, a_2)$ with $(a_1, a_2) \dot{+}(b_1, b_2) \overset{\text{def}}{=} (a_1 + b_2, 0)$, then your vector space $(V, \dot{+})$ over the field $\mathbb{F}$ is isomorphic to the subspace of $\mathbb{F}^2$ spanned by $(1, 0)$, hence isomorphic to the field $\mathbb{F}$ itself.

So, it would be perfectly reasonable to define the scalar product $c(a_1, a_2) = (ca_1, ca_2)$ for scalars $c \in \mathbb{F}$.

On the other hand, given $c\in \mathbb{F}$ and $(a_1, a_2) \in V$, suppose you use the piecewise definition $\star: \mathbb{F} \times V \to V$ for scalar multiplication, where

$$c\star (a_1, a_2) = \begin{cases} (ca_1, 0), & c \neq 1\\ (ca_1, ca_2), & c = 1. \end{cases}$$

(Remember, we require that $1 \star v = v$, for the multiplicative identity $1 \in \mathbb{F}$).

Since we must also have $(c_1c_2) \star v = c_1 \star (c_2 \star v)$ for scalars $c_1, c_2$, then using the above piecewise scalar product, we have

\begin{align*}(a_1, a_2) = 1 \star (a_1, a_2) &= (cc^{-1})\star (a_1, a_2) \\&= c\star(c^{-1} \star (a_1, a_1)) \\ &= c \star(c^{-1}a_1, 0) \\ &= (cc^{-1}a_1, 0) \\ &=(a_1, 0),\end{align*} which is not true for all $(a_1, a_2) \in \mathbb{F}.$

You can perhaps come up with some kind of non-standard scalar product, but if that piecewise definition is the one you were thinking of, it won't work.