Linear bijection between a known and suspected vector space

Question

I have a finite-dimensional real vector space, $V$, and I have written a transformation $T:V\rightarrow W$ that I have confirmed is linear and bijective.

If I knew $W$ to be a vector space, I would conclude $V\cong W$. Alas, I do not know $W$ to be a vector space.

Does the existence of this “would-be” isomorphism $T$ guarantee that $W$ is a vector space? That is, does the existence of this $T$ guarantee that $W$ satisfies the vector space axioms for its definitions of vector addition and scalar multiplication?

Answer 1

In short, yes.

Let

• $(V, +, \,\cdot\,)$ be a vector space over $\Bbb F$,
• $W$ a set equipped with operations
  • $\oplus: W \times W \to W$ and
  • $\odot: \Bbb F \times W \to W$, and
• $T : V \to W$ a bijection satisfying $$T(a \cdot x + b \cdot y) = (a \odot T(x)) \oplus (b \odot T(y))$$ for all $x, y \in V$ and scalars $a, b$ in the field $\Bbb F$ underlying $V$.

We can check each vector space axiom on $W$ directly. For example, the multiplication axiom asks that $a \odot (b \odot w) = (ab) \odot w$ for all $a, b \in \Bbb F$ and $w \in W$. Specializing the above identity to $a = 0$, $y = T^{-1}(w)$ gives $$b \odot w = b \odot T(T^{-1}(w)) = T(b \cdot T^{-1}(w)),$$ so \begin{align} a \odot (b \odot w) &= T(a \cdot T^{-1}(T(b \cdot T^{-1}(w))))\\ &= T(a \cdot (b \cdot T^{-1}(w)))\\ &= T((a b) \cdot T^{-1}(w))\\ &= (ab) \odot T(T^{-1}(w))\\ &= (ab) \odot w , \end{align} as desired. Notice that the third equality just follows from the fact that $(V, +, \,\cdot\,)$ itself satisfies the multiplication axiom, i.e., that $a \cdot (b \cdot y) = (a b) \cdot y$.

More generally, for any vector space $(V, +, \cdot)$ and any set $W$, any bijection $T : V \to W$ determines a vector space structure $(\oplus, \odot)$ on $W$ by rigging the operations $\oplus, \odot$ exactly so that $T$ is an isomorphism. Rearranging the definition of vector space isomorphism gives that these operations must be \begin{align*} w \oplus w' &:= T(T^{-1}(w) + T^{-1}(w')) \\ \lambda \odot w\phantom{'} &:= T(\lambda \cdot T^{-1}(w)) . \end{align*} This maneuver is sometimes called transferring the structure from $V$ to $W$ (via $T$); checking that $(W, \oplus, \odot)$ so defined is a vector space is just a matter of unwinding definitions.