# Operator norm of orthogonal projection

I was assigned the following homework problem:

"Let $$P: H \to H$$ be bounded and linear. Assume it satisfies $$P^2 = P$$ and $$P^\star = P$$. Show $$\|P\| \le 1$$."

This isn't too hard to show: for any $$v\in H$$, $$\| Pv \|^2 = |\langle Pv, Pv \rangle| = | \langle v, Pv \rangle | \le \|v \| \cdot \|Pv\| \implies \frac{\|Pv\|}{\|v\|} \le 1 \implies \|P\|\le 1$$

However, I also noticed the following inequality: $$\|Pv \| = \|P^2 v\| = \|P(Pv)\| \le \|P\| \cdot \|Pv \| \implies \|P \| \ge 1$$

So $$\|P\| = 1$$. But every source I've checked only says $$\|P\| \le 1$$. Is that second inequality true? Why does it fail, if not?

• Consider $P = 0$. That's the exception. May 16, 2014 at 22:27
• Is that the only exception? May 16, 2014 at 22:29

Yes, if $$P^2=P=P^*$$, then $$P$$ is an orthogonal projection to a subspace $$U$$ of $$H$$.
(Prove that $$H={\rm im\,}P\oplus\ker P\$$ and that $$\ {\rm im\,}P\perp\ker P$$.)
The elements of $$U$$ stay fixed under $$P$$, so $$P$$ must have norm $$\ge 1$$ -as you also proved- unless $$U=\{0\}$$ (i.e. $$P=0$$).

• If the projection is not orthogonal, the norm is greater than $1$. May 16, 2014 at 22:49
• Indeed. I corrected the answer. May 18, 2014 at 21:16
• @Berci When $P$ is an orthogonal projection. Please is it true that $\|Px\|=\|x\|$ for all $x\in H$? Aug 7, 2019 at 16:12
• No. It's true for all $x\in im(P)$ only, when we have $Px=x$. For $x\notin im(P)$, we always have $\|Px\|<\|x\|$, basically by the Pythagorean theorem. Aug 7, 2019 at 17:35
The text passage contains a mathematical question and a discussion about its solution, including a proposed answer and a counterexample. It also includes a follow-up question and its answer.
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Given a text passage, determine:
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    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
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# Operator norm of orthogonal projection

I was assigned the following homework problem:

"Let $$P: H \to H$$ be bounded and linear. Assume it satisfies $$P^2 = P$$ and $$P^\star = P$$. Show $$\|P\| \le 1$$."

This isn't too hard to show: for any $$v\in H$$, $$\| Pv \|^2 = |\langle Pv, Pv \rangle| = | \langle v, Pv \rangle | \le \|v \| \cdot \|Pv\| \implies \frac{\|Pv\|}{\|v\|} \le 1 \implies \|P\|\le 1$$

However, I also noticed the following inequality: $$\|Pv \| = \|P^2 v\| = \|P(Pv)\| \le \|P\| \cdot \|Pv \| \implies \|P \| \ge 1$$

So $$\|P\| = 1$$. But every source I've checked only says $$\|P\| \le 1$$. Is that second inequality true? Why does it fail, if not?

• Consider $P = 0$. That's the exception. May 16, 2014 at 22:27
• Is that the only exception? May 16, 2014 at 22:29

Yes, if $$P^2=P=P^*$$, then $$P$$ is an orthogonal projection to a subspace $$U$$ of $$H$$.
(Prove that $$H={\rm im\,}P\oplus\ker P\$$ and that $$\ {\rm im\,}P\perp\ker P$$.)
The elements of $$U$$ stay fixed under $$P$$, so $$P$$ must have norm $$\ge 1$$ -as you also proved- unless $$U=\{0\}$$ (i.e. $$P=0$$).

• If the projection is not orthogonal, the norm is greater than $1$. May 16, 2014 at 22:49
• Indeed. I corrected the answer. May 18, 2014 at 21:16
• @Berci When $P$ is an orthogonal projection. Please is it true that $\|Px\|=\|x\|$ for all $x\in H$? Aug 7, 2019 at 16:12
• No. It's true for all $x\in im(P)$ only, when we have $Px=x$. For $x\notin im(P)$, we always have $\|Px\|<\|x\|$, basically by the Pythagorean theorem. Aug 7, 2019 at 17:35
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The text passage contains a mathematical question and a discussion about its solution, including a proposed answer and a counterexample. It also includes a follow-up question and its answer.

TYPE: SINGLE_QUESTION_WITH_ANSWER