If $p$ and $q$ are orthogonal projections in $\mathbb{R}^n$, then $\Vert p(x)- q(x)\Vert\leq \Vert x\Vert$

Question

I'm trying to adapt a question from Functional Analysis to Linear Algebra. Basically, I'm trying to avoid operator norms.

Let $p$ and $q$ be orthogonal projections in $\mathbb{R}^n$. Then $\Vert p(x)-q(x)\Vert\leq \Vert x\Vert$ for all $x\in\mathbb{R}^n$.

I would like a proof using only the basic properties of orthogonal projections. For example, if $P$ and $Q$ are the ranges of $p$ and $q$ then we can decompose $\mathbb{R}^n=P\oplus P^\perp=Q\oplus Q^\perp=(P\cap Q)\oplus(P\cap Q)^\perp$, so perphaps looking at the coordinates of $x$ on those subspace would give a proof, but I couldn't find a proof using only that.

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Here is a proof using operator norm and a slightly more involved argument that works on any $C^*$-algebra (go to $\mathbb{C}^n=\mathbb{C}\otimes_\mathbb{R}\mathbb{R}^n$ if necessary): We want to prove that $\Vert p-q\Vert\leq 1$ (operator norm).

Let $p^\perp=1-p$ and $q^\perp=1-q$ be the complementary projections of $p$ and $q$. Let $u=p-q$. Then \begin{align*} u&=p-q\\ u^2&=pq^\perp+qp^\perp\\ u^3&=pq^\perp p-qp^\perp q\\ u^4&=pq^\perp pq^\perp+qp^\perp qp^\perp\\ u^5&=pq^\perp pq^\perp p-qp^\perp qp^\perp q\\ \cdots&\cdots\\ u^{2k}&=(pq^\perp)^k+(qp^\perp)^k\\ u^{2k+1}&=(pq^\perp)^kp-(qp^\perp)^kq \end{align*}

All of $p,q,p^\perp$ and $q^\perp$ have norm $\leq 1$, so $\Vert u^n\Vert\leq 2$ for all $n$. Since $u$ is self-adjoint, $\Vert u\Vert^n=\Vert u^n\Vert\leq 2$ for all $n$, thus $\Vert u\Vert\leq 1$.

Answer 1

For convenience, if $T$ is an orthogonal projection map on a (real or complex) Hilbert space $\mathcal{H}$ (such as $\mathbb{R}^n$ with the usual inner product), then $x_T$ and $x_T^\perp$ denote the projections of $x\in\mathbb{R}^n$ onto the image of $T$ and onto the kernel of $T$, respectively. We claim that, for any orthogonal projections $p$ and $q$ on $\mathcal{H}$, $$\big\|(p-q)x\big\|\leq \|x\|\tag{*}$$ for every $x\in\mathcal{H}$. Here, $\langle\_,\_\rangle$ and $\|\_\|$ are, respectively, the inner product on $\mathcal{H}$ and the norm induced by the inner product.

We first note that $$(p-q)^2=p^2-pq-qp+q^2=p-pq-qp+q=p(1-q)+q(1-p)\,,$$ where $1$ here also denotes the identity map on $\mathcal{H}$. For each $x\in\mathcal{H}$, we have $$\big\|(p-q)x\big\|^2=\langle (p-q)x,(p-q)x\rangle = \big\langle x,(p-q)^\dagger(p-q) x\rangle = \big\langle x,(p-q)^2x\big\rangle\,,$$ where $(\_)^\dagger$ is the Hermitian conjugate operator. Thus, $$\begin{align}\big\|(p-q)x\big\|^2&=\Big\langle x,\big(p(1-q)+q(1-p)\big)x\Big\rangle \\&=\big\langle px,(1-q)x\big\rangle +\big\langle qx,(1-p)x\big\rangle\\ &=\langle x_p, x_q^\perp\rangle +\langle x_q, x_p^\perp\rangle=\langle x_p,x_q^\perp\rangle +\overline{\langle x_p^\perp,x_q\rangle}\,.\end{align}$$ Therefore, with the use of the Cauchy-Schwarz Inequality, we get $$\begin{align}\big\|(p-q)x\big\|^2&\leq \Big|\langle x_p,x_q^\perp\rangle\Big|+\Big|\langle x_p^\perp,x_q\rangle\Big| \\ &\leq\|x_p\|\,\|x_q^\perp\|+\|x_p^\perp\|\,\|x_q\| \\&\leq \sqrt{\|x_p\|^2+\|x_p^\perp\|^2}\cdot\sqrt{\|x_q\|^2+\|x_q^\perp\|^2}=\|x\|\cdot\|x\|=\|x\|^2\,. \end{align}$$ The assertion is now verified.

The inequality (*) becomes an equality for a certain $x\in\mathcal{H}$ if and only if $(p+q)x=x$. The equality holds for every $x\in\mathcal{H}$ if and only if $p+q=1$.