Direct sums of vector subspaces I'm struggling to show that if $W_{1},W_{2}$ are subspaces of $V$ such that $W_{1}+W_{2} = V$, then there exists a subspace $W_{3}$ of $W_{2}$ such that $V=W_{1}\oplus W_{3}$.
My idea was to remove from $W_{2}$ vectors in common with $W_{1}$, obtaining $W_{2}'$, in order to make their intersection the zero space. But now I'm having trouble showing that $W_{2}'$ is a subspace of $W_{2}$. Can you please help me?
 A: Choose a base $B_{W_1}=(c_1, c_2, ...)$ for $W_1$ and complete it to a base $B=(c_1, c_2, ..., d_1, d_2, ...)$ of $V$, then consider the projection of $V$ into $W_1$ given by $\psi(c_i)=c_i$, $\psi(d_i)=0$.
Clearly we have $V=\operatorname{im}\psi \bigoplus\ker\psi = W_1 \bigoplus\ker \psi$. $\ker\psi$ is the space you are looking for infact $\ker\psi \subseteq V - W_1 = W_2 - W_1 \subseteq W_2$ (leaving apart the trivial case $W_2 \subseteq W_1$).
A: OK, since someone else has done it using bases, I will show the sleek way that works for finite-dimensional spaces.
Proceed by induction on $\dim(V)$, the result being clear if the dimension is $0$ or $1$.
If $W_2=V$ then one may choose any vector space complement to $V$. Thus just choose any basis for $W_1$ and extend to a basis for $V$. Then the last basis elements are a vector space complement.
If $W_2<V$ then by induction the result holds for $W_2$. Thus one may choose a complement $W_3$ to $W_1\cap W_2$ in $W_2$. Clearly $W_1\cap W_3=0$, and their dimensions sum to $\dim(V)$. Thus $W_3$ is the desired complement.
