Bijection modeling isomorphism of infinite-dimensional vector spaces

Question

Let $T : V \to W$ be an isomorphism of vector spaces with bases $B_V$ and $B_W$, which may be of any cardinality.

Does there exist a bijection $f : B_V \to B_W$ such that, for each $b_V \in B_V$, the coefficient of $f(b_V)$ in $T(b_V)$ is nonzero?

If $V,W$ are finite-dimensional, the answer is yes: $\det(T) \ne 0$, hence some monomial term is nonzero. This exhibits a satisfactory $f$. Alternately, build $f$ inductively by Laplace expansion along a row or column. (Some term $a_{ij} \cdot \text{(complementary minor)}$ is nonzero, and so on.)

Does this hold in general? I have tried using Zorn's lemma, but it seems tricky enough that maybe I'm overlooking a straightforward counterexample.

I can state what I've attempted if there's interest.

(I have tagged this as linear and homological algebra since my desired application is in the latter.)

Answer 1

Yes. Assume $B_V, B_W$ are infinite.

First (merely to simplify notation) reduce to $B_V, B_W$ countable: start with any $v \in B_V$, include the finitely many $w \in B_W$ that are involved in $T(v)$, include the finitely many $v \in B_V$ that are involved in all $T^{-1}(w)$, include the finitely many $w \in B_W$ that are involved in all $T(v)$, etc. Thus $V$ is a direct sum of subspaces each spanned by a countable subset of $B_V$ and mapped isomorphically by $T$ onto a subspace spanned by a countable subset of $B_W$.

Let $B_V = \{{v_1,v_2,...\}}, B_W = \{{w_1,w_2,...\}}$. Write $w_1=a_1T(v_{i_1})+...+a_nT(v_{i_n})$ with all $a_i \neq 0$. Some $T(v_{i_j})$ has non-zero $w_1$-component, say $T(v_{i_1})$.

Claim: $T : \oplus V_{i:i\neq i_1} \simeq W/w_1$ so define $f(v_{i_1})=w_1$ and continue the process inductively, ping-ponging between $T$ and $T^{-1}$ to ensure that $f$ is bijective.

Proof of claim: $T |\oplus V_{i:i\neq i_1}$ is injective because $a_1 \neq 0$; it is onto $w_2$ (for example) because if $w_2=b_1T(v_{k_1})+...+b_mT(v_{k_m})$ and if (say) $k_1=i_1$ then substitute $$b_1T(v_{k_1}) = b_1 a_1^{-1}(w_1-a_2T(v_{i_2})-...-a_nT(v_{i_n})).$$ Note that $f$ has the required property for $T$ and as well $f^{-1}$ has the required property for $T^{-1}$.

Edit: $V_i$ means $Kv_i$.

Edit: Another (similar) way to see the induced smaller map is an isomorphism uses the snake lemma. Let $C_V = \mathrm{span}(v_i : i \ne i_1)$ and $C_W = W/(w_1) = \mathrm{span}(w_i : i \ne 1)$. We have a diagram

$$\begin{array}{ccccccccc}0&& C_V & = & C_V \\&& \downarrow & &\downarrow \tilde{T} \\ (w_1) & \rightarrow & W & \rightarrow & C_W \end{array}$$

which gives from the snake lemma

$$0 \to \ker(\tilde{T}) \to (w_1) \to W/T(C_V) \to \mathrm{coker}(\tilde{T}) \to 0.$$

Note that $W/T(C_V)$ is one-dimensional, spanned by $T(v_{i_1})$, so the middle map is an isomorphism or zero. It is not zero because $w_1 \notin T(C_V)$ (because $a_1 \ne 0$). Thus $\ker(\tilde{T}) = \mathrm{coker}(\tilde T) = 0$.