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Let $P_W,P_V$ be orthogonal projections in Hilbert space $\big(H,\langle\cdot, \cdot\rangle\big)$.

Prove that $P_W+P_V$ is orthogonal projection iff $V\perp W$.

There is hint to consider $(P_W+P_V)^2$ and notice that if $P_V(x)=-x$, then $x=0$

I am aware of solution that does not use this hint, but I want to know how this hint can be used to prove above statement.

mechanodroid
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SekstusEmpiryk
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1 Answers1

1

It is known that $P \in B(H)$ is an orthogonal projection if and only if $P^2 = P^* = P$.

Assume $V \perp W$. Then clearly $P_VP_W = P_WP_V =0$

We have $$(P_W + P_V)^2 = P_W^2 + P_WP_V + P_VP_W + P_V^2 = P_W + P_V$$ so $P_W + P_V$ is idempotent. Also $(P_W + P_V)^* =P_W^* + P_V^* = P_W + P_V$ so $P_W + P_V$ is self-adjoint, so we conclude that it is an orthogonal projection.

Conversely, assume that $P_W + P_V$ is an orthogonal projection. Then $$P_W + P_V = (P_W + P_V)^2 = P_W^2 + P_WP_V + P_VP_W + P_V^2 = P_W + P_WP_V + P_VP_W + P_V$$ which implies $P_WP_V = -P_VP_W$.

Multiplying this from the right by $P_V$ gives $P_WP_V = - P_VP_WP_V$ so

$$P_V(P_WP_V) = -P_WP_V \implies P_WP_V = 0$$

using the hint.

Finally, for $x \in V$ we have

$$ 0 =P_WP_Vx = P_Wx \implies x \perp W$$ so $V \perp W$.

mechanodroid
  • 46,490