Isomorphic vector spaces Let $V$ be a vector space over $\mathbb{R}$ and $V_1$ and $V_2$ be the vector sub-spaces such that $V=V_1\oplus V_2$. If there exists a vector  space $W$ so that $W\simeq V_1$, can one say $W^*\simeq V_2$?
$W^*$ is the dual space of vector space $W$.
Thank you.
 A: It's a good exercise to prove that in finite dimension, a vector space is always isomorphic to its dual (though not canonically.)
With this fact in mind, you can see that if $V_1$ and $V_2$ have different dimension, then $W \simeq V_1 \implies W^* \simeq V_1,$ so that $W^* \not \simeq V_2$. (This criterion holds regardless of whether or not the pullback of $V_1$ and $V_2$ over $V$ is zero, i.e. if their internal direct sum is the same as their (bi)product as vector spaces.)
A: Two relevant facts are:


*

*Two finite-dimensional vector spaces are isomorphic iff they have the same dimension.

*If $W$ is a finite-dimensional vector space, then $W$ and $W^*$ have the same dimension.
Therefore, your statement is true iff $\dim V=2n$ and $\dim V_1 = \dim V_2 =n$.
A: In general the dual $W^\ast$ of a vector space $W$ has nothing to do with the complement of an isomorphic direct copy of $W$ sitting inside some other space $V$.
If the vector spaces are all finite-dimensional and $V = V_1 \oplus V_2$ with $V_1, V_2$ having the same dimension, then $\dim V_1^\ast = \dim V_1 = \dim V_2$, and vector spaces of the same dimension are all isomorphic via the linear extension of any bijection on the basis vectors, so $V_1^\ast \cong V_2$ in this case.
I doubt one can say much more in the general case.
