Symmetric difference of vector subspaces

Question

Let $X$ denote some finite dimensional vector space, and let $P$, $Q$ denote non-trivial commuting projection operators onto subspaces $V$ and $W$. Then,

• $1 - P$ projects onto $V^\perp$.
• $PQ$ projects onto $V \cap W$.
• $P + Q - PQ$ projects onto $V + W$, which is the internal direct sum of $V$ and $W$.

One can make an analogy to propositional logic here where $V^\perp$ is like a "not" subspace, $V \cap W$ is like an "and" subspace, and $V + W$ is like an "or" subspace. One can therefore construct a projector $P + Q - 2PQ$ onto the "xor" or "symmetric difference" subspace. My question is: is it possible to write the space $P + Q - 2PQ$ projects onto in terms of $V, W, ^\perp, \cap, +$? The obvious guess here is that $P + Q - 2PQ$ projects onto $W = (V + W) \setminus (V \cap W)$, but of course this is not a vector space since $0 \not \in W$.

Answer 1

Assume we're working in an inner product space $V$.

Relative Complement. If we want such an operation, it ought to be "complementary" to $A\cap B$ within $A$, just as for sets. The obvious choice is the relative orthogonal complement:

$$ A\setminus B ~:=~ (A\cap B)^{\perp A} ~=~ A\cap(A\cap B)^\perp. $$

E.g. if a pair of $2$D planes $A,B$ intersect in a $1$D subspace, $A\setminus B$ would be the axis in $A$ which is perpendicular to the shared axis. This operation satisfies a number of properties:

• $A\setminus B= A\cap B^\perp\,$ if $\,B\le A$
• $A\setminus B=\varnothing\,$ if $\,A\le B$
• $A\setminus B=A\,$ if $\,A\cap B=0$
• $(A\setminus B)\le A$
• $(A\setminus B)\cap B=0$
• $A=(A\setminus B)\oplus(A\cap B)$ is an orthogonal direct sum

Symmetric Difference. Then we may define a "symmetric difference" operation by

$$ A \,\triangle\, B ~=~ (A+B)\setminus(A\cap B) ~=~ (A\setminus B)+(B\setminus A). $$

Note the above sum is direct, but generally not orthogonal. For example, if $A,B$ are a pair of $2$D planes intersecting in an axis, $A\,\triangle\,B$ is the $2$D plane which meets both of them perpendicularly (i.e. at a right dihedral angle) but is still contained in the same $3$D space they span. (Note the angle between the $1$D axes $A\setminus B$ and $B\setminus A$ is the dihedral angle between $A$ and $B$.)

This "symmetric difference" satisfies a number of properties:

• $A\,\triangle\,0=A$
• $A\,\triangle\,A=0$
• $A\,\triangle\,V=A^\perp$
• $A\,\triangle\,B=B\,\triangle\,A$
• $A+B=(A\,\triangle\,B)\oplus(A\cap B)$

The direct sum above is orthogonal.

Non-Associativity. While commutative, $\triangle$ is not associative, unlike its set-theory counterpart:

$$ (A\,\triangle\,B)\,\triangle\,C ~\ne~ A\,\triangle\,(B\,\triangle\,C). $$

To see this, consider $B'=A\,\triangle\,(A\,\triangle\, B)$ when $A,B$ are once again a pair of $2$D planes with a $1$D intersection: then $B'$ is the $2$D plane, contained in the same $3$D span as $A$ and $B$, which meets $A$ at a right dihedral angle. That is, $B'$ is like a "straightening" of $B$ relative to $A$. Indeed,

$$ A\,\triangle\,(A\,\triangle\, B) ~=~ ((A+B)\setminus A)\oplus (A\cap B) $$

replaces the $B\setminus A$ in $B=(B\setminus A)\oplus(A\cap B)$ with the image of $B\setminus A$ under orthogonal rejection from $A$, i.e. with $(A+B)\setminus A$. The result can be different from $B=(A\,\triangle\,A)\,\triangle\,B$.