can we say that $P_{X\oplus Y}=P_X+P_Y$?

Question

if $X$ and $Y$ are two closed subspaces in H then can we say that $P_{X\oplus Y}=P_X+P_Y$? Here $P_X$ is orthogonal projection on X.

Answer 1

Yes if $X\perp Y$, no if they are not orthogonal.

Concretely, $X\perp Y$ if and only if $P_X+P_Y$ is a projection. When this happens, $P_{X\oplus Y}=P_X+P_Y$.

Indeed, suppose first that $X\perp Y$. Then $H=X\oplus Y\oplus Z$, and $$P_{X\oplus Y}(x+y+z)=x+y=P_X(x+y+z)+P_Y(x+y+z).$$ So $P_{X\oplus Y}=P_X+P_Y$.

Conversely, if $P_X+P_Y$ is a projection, then $$P_X+P_Y=(P_X+P_Y)^2=P_X+P_Y+P_XP_Y+P_YP_X.$$ So $P_XP_Y+P_YP_X=0$. Multiplying by $I-P_Y$ on the left, we get $(I-P_Y)P_XP_Y=0$. As $P_X^*=P_X$, $P_Y^*=P_Y$, taking adjoints we get $P_YP_X(I-P_Y)=0$. Thus $$ P_X=P_YP_XP_Y+(I-P_Y)P_XP_Y+P_YP_X(I-P_Y)+(I-P_Y)P_X(I-P_Y)\\ =P_YP_XP_Y+(I-P_Y)P_X(I-P_Y) $$ Now multiplying by $P_Y$ on the left we get $$P_YP_X=P_YP_XP_Y.$$ As the term of the right is selfadjoint, we deduce (taking adjoints) that $P_YP_X=P_XP_Y$. Then $$ 0=P_XP_Y+P_YP_X=2P_YP_X; $$ thus $P_XP_Y=0$, and so $X\perp Y$.