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Assume that $p$ and $q$ are (orthogonal) projections on Hilbert space $\mathcal{H}$. I want to prove: $pq=0$ iff $p+q\leq1$

I had the following in mind: Assume $pq=0$. Then $qp=0$, hence $p+q$ is a projection. One has the theorem that if e and f are two projections, then $e\leq f$ iff $ef=f$. But if we take $e=p+q$ and $f=1$, then $p+q\leq p+q$ which is true. Hence $p+q<1$.

For the other direction I thought about an estimate of $||pqx||$, but I don't know how to go further. Anyone who can help me with this? Furthermore, if there are any corrections about the other directions proof, please let me know.

Thanks

glS
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Isidora Bernardita
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2 Answers2

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  • The assumption $p+q\le 1$ is equivalent to $$ \forall x\in\mathcal{H},\qquad<(p+q)x,x>\le \Vert x\Vert^2$$ or equivalently $$ \forall x\in\mathcal{H},\qquad \Vert p (x)\Vert^2+\Vert q(x)\Vert^2\le \Vert x\Vert^2$$ Applying this to $x=q(y)$ we get $$ \forall y\in\mathcal{H},\qquad \Vert p (q(y))\Vert^2+\Vert q(y)\Vert^2\le \Vert q(y)\Vert^2$$ That is $pq(y)=0$ for all $y\in\mathcal{H}$ This proves the missing direction.
  • For the first direction we need to show that $e=p+q$ is an orthogonal projection which is easy since clearly $e^*=e$ and as you have shown $e^2=e$ because $pq=qp=0$. Now every orthogonal projection $e$ satisfies $e\le 1$. Indeed since for every $x$ the vectors $e(x)$ and $(1-e)(x)$ are orthogonal we have $$\Vert x\Vert^2= \Vert e(x)\Vert^2+\Vert (1-e)(x)\Vert^2\ge \Vert e(x)\Vert^2$$ that is $<e(x),x>\,=\,<e(x),e(x)>\le <x,x>$ for every $x\in\mathcal{H}$.
Omran Kouba
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Here is another way to prove this, similar to the method used for example in this other answer:

  1. Suppose $pq=0$. Then $(p+q)^2=p+q$, hence $P\equiv p+q$ is a projection as well. Projections can only have $0$ and $1$ as eigenvalues (e.g. because their minimal polynomial is $x\mapsto x(1-x)$), thus $\langle x,Px\rangle\le1$ for all $x$, that is, $P\le I$.

  2. Suppose now $p+q\le I$. Then $p\le I- q$. Multiplying by $q$ on left and right, this becomes $qpq\le q(I-q)q=0$. But $qpq=qppq=(qp)(qp)^\dagger\ge0$, hence $qpq=0$ and $qp=pq=0$.

glS
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