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.)



Bijection modeling isomorphism of infinite-dimensional vector spaces
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 $\ker(\tilde{T}) = \mathrm{coker}(\tilde T) = 0$.


