# Cancelling a direct product of a subspace in isomorphic ambient spaces?

I read today that for a vector space $V$ and any subspace $S$, all complements of $S$ in $V$ are isomorphic to $V/S$, and thus to each other.

I want to ask, is there a case where $V\cong W$ are isomorphic vector spaces, where $V=S\oplus A$ and $W=S\oplus B$, but $A$ and $B$ are not isomorphic? So complements in different vector spaces, albeit isomorphic ones, need not be isomorphic themselves? Thanks.

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Yes, there are such cases. If $S$ is infinite dimensional, you can take $V=S\oplus S$ and $W=S\oplus\{0\}$. Or, $V=S\oplus(\text{a finite dimensional space})$ and $W=S\oplus(\text{a finite dimensional space with different dimension})$.

If $S$ is finite dimensional, then there are no such cases.

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Thanks Jonas Meyer. – Noomi Holloway Dec 21 '12 at 3:48

The following works if the bases of $V$ and $W$ are finite.

Given an isomorphism $\phi:S\rightarrow S$, we can extend to an isomorphism $\overline{\phi}:V\rightarrow W$ by mapping the completion of the basis of $S$ in $V$ to the completion of the basis of $S$ in $W$. Then $\overline{\phi}\left|_{A}\right.$ is an isomorphism between $A$ and $B$.

I don't know about infinite dimensions.

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@JonasMeyer Yeah, I realize that. Thanks anyway for pointing out the error. – Samuel Handwich Dec 20 '12 at 8:10