There are some things about projections that, it turned out, I was getting all wrong. Specifically, it concerns the norms of projections. All right, I shall confess: I never realized the norm could be greater than one.

## Recap

A linear operator $$P: X\to X$$ is called a projection if $$P^2 = P$$. Suppose $$X$$ is a normed v.s. Then we can introduce the "operator norm": $\|A\|_{\mathrm{op}} = \inf \{C\in\mathbb{R};\ \|Ax\|\leq C\|x\|\ \text{for all}\ x\in X\}$ for those operators $$A$$ for which it's finite. Here $$\|\cdot\|:X\to\mathbb{R}$$ is the norm in $$X$$. That value happens to equal $\|A\|_{\mathrm{op}} = \sup_{x\neq 0} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\|=1} \|Ax\|.$ Such operators form a linear subspace of the space of all operators on $$X$$. Moreover they form a Banach Algebra with the usual multiplication (composition) of operators and the submultiplicative norm $$\|\cdot\|_{\mathrm{op}}$$: $\|AB\|_{\mathrm{op}} \leq \|A\|_{\mathrm{op}}\|B\|_{\mathrm{op}}$

## Wrong intuition

The (wrong) first intuition is that for a non-zero projection there will be $$\|P\|=1$$. Here's where this deception comes from. I'd erroneously think of orthogonal projections and expect Px = \left\{\begin{aligned} & x,\ \text{for}\ x\in X_1,\\ & 0,\ \text{otherwise}.\end{aligned}\right. Then obviously $$\|P\| = \sup_{\|x\|=1} \|x\| = 1.$$ Yet the projection needs not be of the form above --- it may be oblique.

Here's an example AG Baskakov used to demonstrate this phenomenon to me --- his ignorant student. Consider $$X = \mathbb{R}^n$$ and an operator $$P$$ defined by the formula: $Px = (x, a) b,$ for some fixed $$a, b \in X$$. Then $$P^2 x = Py = (y, a) b = (x, a) (b, a) b,$$ where $$y = Px = (x, a) b$$. Thus chose we such $$a, b$$ that $$(a, b) = 1$$ the operator $$p$$ would be a projection. Now to its norm: $\sup_{\|x\|=1} \|Px\| = \sup_{\|x\|=1} |(x, a)|\|b\| = (\frac{1}{\sqrt{a}}a, a)\|b\| = \|a\|\|b\|.$ Could we pick $$a$$ and $$b$$ to satisfy $$(a, b) = \|a\|\|b\|\cos\phi = 1$$ and $$\|a\|\|b\| > 1$$? Most certainly, sir.

## Fix

But let's just see what can be told from the definition. Since $$P^2=P$$ and since the norm is submultiplicative it's clear that: $\|P\|_{\mathrm{op}} = \|P^2\|_{\mathrm{op}} \leq \|P\|_{\mathrm{op}}^2.$ This in turn implies $1\leq \|P\|_{\mathrm{op}}$