Mathematics Asked by Pedro Mariz on March 5, 2021
Assuming the Axiom of Choice, and therefore assuming that every vector space has a Hamel basis, it isn’t too hard to show that if $U$ and $V$ are vector spaces with the same dimension (i.e. their bases have the same cardinality), then $Vcong U$. However, I’m not so sure how to deal the converse implication (assuming it’s true in general in the first place):
Let $mathcal B_1$ be a basis for $V$ and $mathcal B_2$ a basis for $U$. If $Vcong U$ then there exists a linear map $phi: Vto U$ that is bijective, and so particularly, $text{Im}(phi) = U = langle mathcal B_2rangle$.
I’m not sure how to conclude that $phi(mathcal B_1)=mathcal B_2$. I thought of using the First Isomorphism Theorem ($V/text{ker}(phi)cong text{Im}(phi)$) to mimic the Rank-Nullity Theorem being used for the finite-dimensional case, but it doesn’t really help in this scenario.
Suppose that $mathcal B_1 = (v_alpha : alpha in I)$ is a basis of $V$ and that $phi: V to U$ is a bijective linear map. We now consider $mathcal B_2 = phi(mathcal B_1) = (phi(v_{alpha}): alpha in I)$.
Claim: $phi(mathcal B_1)$ spans $U$.
Proof: Consider an arbitrary $u in U$. $phi$ is surjective, so there exists a $v in V$ with $phi(v) = u$. Because $mathcal B_1$ is a (Hamel) basis, there exist elements $v_{alpha_1}, dots, v_{alpha_n}$ and coefficients $c_1,dots,c_n$ such that $$ v = sum_{i=1}^n c_i v_{alpha_i} implies u = phi(v) = sum_{i=1}^n c_i phi(v_{alpha_i}). $$
Claim: $phi(mathcal B_1)$ is linearly independent.
Proof: Suppose that $v_{alpha_1},dots,v_{alpha_n}$ and coefficients $c_1,dots,c_n$ are such that $sum_{i=1}^n c_i phi(v_{alpha_i}) = 0$. It follows that $phi left(sum_{i=1}^n c_i v_{alpha_i} right) = 0$. Because $phi$ is injective, it must holds that $sum_{i=1}^n c_i v_{alpha_i} = 0$. Because $mathcal B_1$ is a basis, this can only hold if $c_{alpha_1} = cdots = c_{alpha_n} = 0$.
So, $phi(mathcal B_1)$ is indeed linearly independent.
Because $phi(mathcal B_1)$ is a linearly independent spanning set, it is a basis.
Correct answer by Ben Grossmann on March 5, 2021
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