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Suppose we have Hermitian operators $P_1$ and $P_2$ on Hilbert space $\mathbb{H} \cong \mathbb{C}^n$, such that: $P_1^2=P_1$, $P_2^2=P_2$, $\{P_1, P_2\} = 0$. Let $x$ be a unit $L_2$-norm vector.

Show that $(x^TP_1x)^2 + (x^TP_2x)^2 \leq 1$

I can only see how one can show that the same inequality holds for $RHS=2$. That can be easily seen because of the fact that a projection of a vector is always "shorter" than the vector. But I'm confused about the $RHS=1$ case.

I think I should somehow use the restriction on the anticommutator, but I can't figure that out. It is easy to prove that in this case $P_1+P_2$ is also a projector but I can't see how this helps here.

mechanodroid
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Aleksandr Berezutskii
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2 Answers2

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In fact, the correct statement should be$$ x^TP_1x + x^TP_2x \leq 1,\quad\cdots(*). $$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let $$ x= P_1x + P_2 x + P_3x,$$ where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $i\neq j$. Thus, we have $$ 1=x^Tx =\sum_{i=1}^3 x^TP_ix +\sum_{i\neq j}x^TP_iP_jx= \sum_{i=1}^3 x^TP_ix \geq x^TP_1x + x^TP_2x, $$ as desired.

$\textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.

Assume to the contrary that $PQ\neq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pv\neq 0$. By the assumption that $PQ=-QP$, we have $$ Pv = -QPv. $$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.

Myunghyun Song
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The condition $P_1P_2 + P_2P_1 = 0$ holds if and only if $P_1$ and $P_2$ project onto orthogonal subspaces. Also it is equivalent to $P_1 + P_2$ being an orthogonal projection.

Notice that for an orthogonal projection $P$ and $x \in \Bbb{H}$ we have $$\|Px\|^2 = \langle Px,Px\rangle = \langle P^*Px,x\rangle = \langle P^2x,x\rangle = \langle Px,x\rangle$$

Hence for $x \in \Bbb{H}$, $\|x\| = 1$ we have

\begin{align} LHS &= \langle P_1x,x\rangle^2 + \langle P_2x,x\rangle^2 \\ &= \|P_1x\|^4 + \|P_2x\|^4 \\ &\le \|P_1x\|^4 + 2\|P_1x\|^2\|P_2x\|^2 + \|P_2x\|^4 \\ &= (\|P_1x\|^2+\|P_2x\|^2)^2 \end{align}

Since $P_1x \perp P_2x$ we have

$$LHS \le (\|P_1x\|^2+\|P_2x\|^2)^2 = \|P_1x + P_2x\|^4 = \|(P_1+P_2)x\|^4 \le \|x\|^4 = 1$$

because $P_1 + P_2$ is an orthogonal projection and hence of norm $\le 1$.

mechanodroid
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