# Projections

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

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