Let $K$ be a closed convex cone in $\mathbb{R}^n$.
Show that $K = (K \cap -K) + (K \cap (K \cap -K)^{\perp})$
- The above holds if $K = -K$, because $K$ is then a subspace.
- Similarly, it holds if $K$ is pointed (i.e. $K \cap -K = \{0\}$).
However I am not able to prove it in the general case of the statement. Any suggestion? Thanks.
Note that $K \cap -K$ and $K \cap (K \cap -K)^\perp$ are both subsets of $K$, and since convex cones are closed under addition, $$K \supseteq (K \cap -K) + (K \cap (K \cap -K)^\perp).$$ We now need to prove the reverse inclusion. That is, we need to be able to take an arbitrary $x \in K$ and write it as a sum of two vectors $y \in K \cap -K$ and $z \in K \cap (K \cap -K)^\perp$. The question is, how can we figure out what these vectors are?
Well, what we seem to have here is actually an orthogonal decomposition. Note that $y$ and $z$ are orthogonal to each other, where $y$ lies on the subspace $K \cap -K$, and $z$ lies in the perpendicular complement of this subspace. So, it would seem appropriate to let $y = \operatorname{proj}_{K \cap -K} x$, and let $z = x - y$. Clearly $x = y + z$, clearly $y \in K \cap -K$, and clearly $z \in (K \cap -K)^\perp$. The only missing piece of the puzzle is to show that $z \in K$.
But this is simple enough. Note that $y \in -K$, so $-y \in K$. As such, $z = x + (-y)$ is the sum of two vectors in $K$, and hence is in $K$. Thus $z \in K \cap (K \cap -K)^\perp$ as required, and hence $$x \in (K \cap -K) + (K \cap (K \cap -K)^\perp).$$