Isomorphism of X to itself.

Question

Let $(X, +, \bullet)$ and $(X,\tilde{+}, \tilde{\bullet})$ be vector spaces over $\mathbb{R}$, and dim$(X, +, \bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, \bullet)$ is isomorphic to $(X,\tilde{+}, \tilde{\bullet})$.

Answer 1

Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:

$$(X, +, \bullet)\cong(X,\tilde{+}, \tilde{\bullet})$$

if and only if $$\dim (X,\tilde{+}, \tilde{\bullet}) = n$$

Answer 2

The fact that the set is the same is a distraction. Choose a basis $(v_1, \dots,v_n)$ for $(X,+,\cdot)$. Choose a basis $w_1, \dots w_n$ for $(X, \tilde{+}, \tilde{\cdot})$. Define $T:X \to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(\alpha_1 \cdot v_i+ \alpha_2 \cdot v_j)=\alpha_1 \tilde{\cdot}T(v_i)\tilde{+}\alpha_2\tilde{\cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, \tilde{+}, \tilde{\cdot})$. By the rank nullity theorem it will be invective. (Check that this defines a linear map)

(This answer assumes that the dimensions are the same)