I want to prove this proposition:

Let $H$ be an Hilbert space and let $M,N$ closed subspaces of $H$. Let$P_M$ and $P_N$ orthogonal projections with range(M) and range(N) respectively. Prove that $P_M + P_N = P_{M+N}$ if and only if $M \perp N$.

According to what I have studied in Regression analysis and linear algebra, I understand this proposition perfectly and my geometrical imagination capacity says that it is something obvious, but I don't know how to prove it, nethier how to start the proof (For me, this is just by definition of projection).


Suppose first that $M\perp N$. Then, for any $x,y\in H$, $$ \langle P_MP_Nx,y\rangle=\langle P_Nx,P_My\rangle=0. $$ As we can do this for all $x,y$, we get that $P_MP_N=0$. Then $$ (P_M+P_N)^2=P_M^2+P_N^2=P_M+P_N $$ and $P_N+P_M$ is a projection. Given $x+y\in M+N$ with $x\in M$ and $y\in N$, $$(P_M+P_N)(x+y)=P_Mx+P_Nx+P_My+P_Ny=x+y,$$since $P_Mx=x$, $P_nx=0$, $P_My=0$, $P_Ny=y$. If $z\in (M+N)^\perp$, then $$\langle(P_M+P_N)z,w\rangle=\langle z,(P_M+P_N)w\rangle=0.$$ So $P_M+P_N=P_{M+N}$.

Conversely, if $P_M+P_N=P_{M+N}$, then $$ P_M+P_N=P_{M+N}=P_{M+N}^2=P_M+P_N+P_NP_M+P_MP_N. $$ Then $$\tag{*}P_MP_N+P_NP_M=0.$$ Multiply on the left by $I-P_M$, and we get $(I-P_M)P_NP_M=0$. This is $P_MP_NP_M=P_NP_M$. Then $$ P_MP_N=(P_NP_M)^*=(P_MP_NP_M)^*=P_MP_NP_M=P_NP_M. $$ Then $P_NP_M=P_MP_N$ and now $(*)$ becomes $P_MP_N=0$. If we take $x\in M$ and $y\in N$, $$ \langle x,y\rangle=\langle P_Mx,P_Ny\rangle=\langle P_NP_Mx,y\rangle=0. $$ So $M\perp N$.

  • $\begingroup$ I'm trying to understand the proof now :) Thanks you for your answer. $\endgroup$ – Applicable Math Oct 8 '16 at 16:29
  • $\begingroup$ Sorry for commenting on such an old answer. I was wondering if when we assume $M\perp N$, is it obvious that $M+N$ is closed in the first place, so that $P_{M+N}$ is well defined? Is this implied somewhere within your proof, or would one need to prove this separately? $\endgroup$ – lkjhgfdsa Mar 13 '18 at 12:11
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    $\begingroup$ It is more or less obvious. If $z_n=x_n+y_n$ is Cauchy, from $\|z_n\|^2=\|x_n\|^2+\|y_n\|^2$ you get that each command is Cauchy. $\endgroup$ – Martin Argerami Mar 13 '18 at 14:03

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