I am curious to know if theres a way of proving that the n-th dimensional spaces U and W have the same basis. That is, showing that a vector space as a subspace of itself is itself, by showing they have the same basis and same dimension.
Given U is an n-th dimensional subspace of an n-th dimensional vector space W. Im wondering if this information implies enough vector space properties to show they have the same basis vector(s).
That is enough. Let $(e_i)_{i\leq n}$ be a basis of $U$ then each $e_i$ is in $W$ since $U\subset W$. Then $(e_i)_{i\leq n}$ is also a basis of $W$ because they remain linearly independent (the equation for linear independence remains the same whatever the vector space) and there are as many as the dimension of $W$.
You then have: $U=\text{Span} (e_1,...,e_n)=W$.