Intersection of three vector subspaces We say that two vector subspaces $U,W\subset V$ intersect transversally (written $U\pitchfork W$) if $U+W=V$.
Let $V_1,V_2,V_3$ be vector subspaces of $V$ where $\dim V=n$. We say they intersect normally if $V_i\pitchfork(V_k\cap V_l)$
 for $i\ne k,l$. Prove that $V_1,V_2,V_3$ intersect normally if  the following condition holds:
$\operatorname{codim}(V_1\cap V_2\cap V_3)=\operatorname{codim}
V_1+\operatorname{codim} V_2+\operatorname{codim} V_3,$ 
where $\operatorname{codim} V_i=n-\dim V_i$.
How to approach this problem? I tried to order dimensions of $V_i$ but I got that the LHS is $\ge n-\min (\dim V_i,i=1,2,3)$ and the RHS is $\le 3( n-\min (\dim V_i,i=1,2,3))$ which does not lead anywhere. What is the right way to do this?
 A: First thing to note is that $V_1 \pitchfork V_2$, $V_2 \pitchfork V_3$ and $V_3 \pitchfork V_1$. Why? Because
$$V = (V_1 \cap V_2) + V_3 \le V_2 + V_3 \le V.$$
The rest is just using the identity
$$\operatorname{dim}(U + W) = \operatorname{dim}(U) + \operatorname{dim}(W) - \operatorname{dim}(U \cap W),$$
or, turning it into a statement about codimension,
$$\operatorname{codim}(U + W) = \operatorname{codim}(U) + \operatorname{codim}(W) - \operatorname{codim}(U \cap W).$$
In particular, if we consider $U = V_1$ and $W = V_2 \cap V_3$, we get
\begin{align*}
\operatorname{codim}(V_1 \cap V_2 \cap V_3) &= \operatorname{codim}(V_1) + \operatorname{codim}(V_2 \cap V_3) - \operatorname{codim}(V_1 + (V_2 \cap V_3)) \\
&= \operatorname{codim}(V_1) + \operatorname{codim}(V_2 \cap V_3) - 0 \\
&= \operatorname{codim}(V_1) + \operatorname{codim}(V_2) + \operatorname{codim}(V_3) - \operatorname{codim}(V_2 + V_3) \\
&= \operatorname{codim}(V_1) + \operatorname{codim}(V_2) + \operatorname{codim}(V_3) - 0.
\end{align*}

EDIT: I proved the converse accidentally. You can recover the original without too much hassle though. Just take the equality as proven (but not stated) above:
\begin{align*}\operatorname{codim}(V_1 \cap V_2 \cap V_3) &= \operatorname{codim}(V_1) + \operatorname{codim}(V_2) + \operatorname{codim}(V_3) \\
&- \operatorname{codim}(V_2 + V_3) - \operatorname{codim}(V_1 + (V_2 \cap V_3))
\end{align*}
If we assume
$$\operatorname{codim}(V_1 \cap V_2 \cap V_3) = \operatorname{codim}(V_1) + \operatorname{codim}(V_2) + \operatorname{codim}(V_3),$$
then
$$0 = \operatorname{codim}(V_2 + V_3) + \operatorname{codim}(V_1 + (V_2 \cap V_3)) \ge \operatorname{codim}(V_1 + (V_2 \cap V_3)) \ge 0.$$
Thus
$$\operatorname{codim}(V_1 + (V_2 \cap V_3)) = 0 \implies V_1 + (V_2 \cap V_3) = V.$$
The other cases follow similarly.
