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
\[ 1\leq \|P\|_{\mathrm{op}} \]

## Features of orthogonal projections