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We have the following theorem :

Let $P:H\to H$ be a bounded linear map on a complex vector space satisfying $P^2=P$. Then the following are equivalent: $(1)$ $P$ is self-adjoint $(2)$ $P$ is normal $(3)$ $x-Px$ is orthogonal to $Px$ for every $x\in H$. Also, if these conditions hold then $P$ is the orthogonal projection onto its image.

Now I was trying to find a linear bounded map that satisfies none of these, but failed to do so. What would be an easy to see example of such a map that is not orthogonal projection? (The reason I stated the theorem is because now we know that it is equivalent to find such a map that is not self-adjoint)

User Not Found
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2 Answers2

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Take $$P=\begin {bmatrix} 1&1\\0&0\end {bmatrix}. $$ By "putting more zeroes" you can generalize this idea to any $H $ with $\dim h\geq2$.

Martin Argerami
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  • By bounded I mean $||Pv||\le||v||$. That is the standard definition right? Why is this bounded? – User Not Found Dec 18 '17 at 08:28
  • @UserNotFound In a finite dimensional vector space, every linear operator is bounded – Stefan Dec 18 '17 at 10:04
  • @Stefan Is it easy to see? My knowledge in these fields is limited. – User Not Found Dec 18 '17 at 10:27
  • @UserNotFound: Boundedness is equivalent to continuity. Do you know that every linear operator on a finite dimensional vector space is continuous? – Stefan Dec 18 '17 at 13:12
  • @Stefan Yes I know that....ok let me just ask the question and you say yes or no: "$P$ is an operator on a complex vector space satisfying $P^2 = P$ and $||P v|| \le ||v||$ for all $v$". I wanted to show that $P$ need not be an orthogonal projection. Are we on the same page? – User Not Found Dec 18 '17 at 13:27
  • @UserNotFound: if $P^2=P$ and $|P|\leq1$, then $P$ is an orthogonal projection.See here. – Martin Argerami Dec 18 '17 at 13:37
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    @UserNotFound: the answers and comments at Math.SE are not the right place for a class on Functional Analysis, and I think you need to learn chapter 1, section 1 of any text on operators on a normed space. The assertion "$|Pv|\leq|v|$ for all $v$" is exactly the same as "$|P|\leq1$". And it is more than "bounded", which would be "$|Pv|\leq c,|v|$ for a fixed $c$ and all $v$". – Martin Argerami Dec 18 '17 at 13:50
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Try an oblique projector (from Carl Meyer's Matrix Analysis and Linear Algebra)

$$P := \begin{pmatrix} -2 & -2 & -1 \\ 3 & 3 & 1\\ 0 & 0 & 1\end{pmatrix}$$ is an idempotent (i.e., $P^2 = P$) but clearly $P$ is not self-adjoint.

max_zorn
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