How to show that you can find two subspaces that don't intersect Suppose $V, V'$ are subspaces of dimension $d$ of a vector space $X$. Then there is a subspace $W$ of $X$ of codimension $d$ such that $W \cap V = W \cap V' = { 0 }$. 
This can be proved by choosing an explicit basis for $X$ which contains a basis for $V$ and a basis for $V'$ and a basis for $V \cap V'$.  On the other hand, there should be a nice way to do this without choosing a basis. Can anyone explain this?
Disclosure: This came up when I was doing a homework problem. However, I'm just going to use the non-basis-free approach when I write up my answer.
 A: Suppose first that $V+ V' = X$.  Then the natural map $X/(V\cap V') \to X/V \times X/V'$
is an isomorphism.  Choose a subspace of the target of this isomorphism
that projects isomorphically onto each factor (i.e. the graph of an isomorphism between the two factors; such an isomorphism exists since the two factors have the same dimension).
Its preimage under the natural map is a subspace of $X/(V\cap V')$ which
meets each of $V$ and $V'$ trivially.  Now choose any  subspace $W$
of $X$ that projects isomorphically onto this preimage; this is then a subspace of
$X$ that maps isomorphically onto each of $X/V$ and $X/V'$, and hence is a codimension $d$
subspace with the desired property.
In general (i.e. if $V + V' \neq X$) then the above gives a codimension $d$ subspace $W'$ of 
$V + V'$ meeting $V$ and $V'$ trivially.   Choose $W''$ to be any subspace of $X$ which
maps isomorphically onto $X/(V + V')$.  The sum $W' + W''$ is then a codimension $d$ subspace
of $X$ meeting $V$ and $V'$ trivially.
A: So let $G$ be the space of all co-dimension d subspaces. ($G$ is
naturally a manifold). I claim that if $W$ is any dimension $d$ subspace,
the set of $X$ s.t. $X \cap W = 0$ is an open, dense subset of $G$.
Open is easy. If you have $X_i$ each with non-trivial intersection with
W and $X_i \to X$ then you can pick lines $L_i$ in $W$ intersect $X_i$ and find
some limit (perhaps of a subsequence) so that $L_i \to L$. Then $L$ is in
$W \cap X$.
To show dense is also not hard. You need to show that if $X$ intersects
$W$, then you can change $X$ by a little so that it no longer does. This
really isn't hard but it is annoying to do without ever picking a
basis of anything. How about this. You can think of $X$ as the image
of a map $Y \to Z$ ($Y$ is a space of dim $n-d$, $Z$ is your big space). $X$
intersects $W$ trivially iff the map $Y \to Z/W$ is an injection. Pick
some $U$ so that $Z = U + V$. Then we can think of our map as $Y \to U+V$,
and we want the map $Y \to U$ to be injective. But $Y$ and  $U$ have the same
dimension and it is easy to modify a map between such spaces by
epsilon to make it a bijection (for example add a small multiple of
some particular bijection).
