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Letting $F$ be a field and $V$ a finite-dimensional vector space over $F$, we have $U_i,\ldots,U_k$ distinct proper subspaces of $V$. I'm asked to provide a condition on the $U_i$ such that there exists a subspace $W$ such that $U_i\oplus W = V$ for all $i$, where $\oplus$ denotes the internal direct sum. My claim is that $\dim(U_i)=\dim(U_j)$ for all $i$ and $j$ is the necessary condition. Moreover I want to know that this necessary condition is also sufficient, namely, there does exist such a unique $W$ in the event that $F$ is infinite.

Showing the necessary-ness seems reasonable in that each $U_i$ should be isomorphic to $V/W$ if such a $U$ exists, but actually constructing such a $W$ for the case that $F$ is infinite is tripping me up. I'm assuming there's a clever way of extending the bases of $U_i$, but I haven't really been able to make that precise.

It may be that it's closely related to a previous question I posed here

As always, any and all help is very much appreciated.

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We assume $F$ is infinite. It is known that $V$ is not a union of any finite set of proper subspaces (I believe this is what the linked question is about). Choose $w$ in $V$ not in the union of the $U_i$, let $W$ be the subspace generated by $w$. If $U_i\oplus W=V$ for all $i$, you are done. If not, then for all $i$, $U_i\oplus W$ is a proper subspace of $V$, and you can choose $w'$ in $V$ but not in the union of the $U_i\oplus W$. Let $W'$ be the span of $w$ and $w'$. Iterate until you run out of dimensions --- then you'll have the space you're looking for.

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    Bit of a late follow up, but if I wanted to make this 100% precise, I'd use induction yes? But what am I inducting on? Is it on $dim(V)-dim(W_i)$? Or rather, $dim(W_i)$, since $dim(V)$ is held fixed.2012-10-23