TITLE: How to show that you can find two subspaces that don't intersect
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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.