It's well known that an orthogonal projection satisfies $\|Px\|\leq\|x\|$.
Does this property hold for any general projection operator $P$, which is defined by $P^2=P$?
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possible duplicate of An example of an unbounded non-orthogonal projection in a Hilbert space – Prism Jul 09 '15 at 03:32
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The property does not hold. For example, consider $$ P = \pmatrix{1&0\\\alpha &0} $$ for some $\alpha \neq 0$. Note that $P^2 = P$, but $$ \left\| \pmatrix{1&0\\\alpha &0} \pmatrix{1\\0} \right\| = \left\| \pmatrix{1\\ \alpha} \right\| \geq \left\| \pmatrix{1\\0} \right\| $$
Ben Grossmann
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1In fact, it can be shown that, if $P^2 = P$, then $|Px| \leq |x|$ for every $x$ if and only if $P$ is an orthogonal projection. – Ben Grossmann Jul 09 '15 at 03:54
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Thank you very much for a clear counterexample. Could you please explain to me why this property hold if and only if $P$ is an orthogonal projection? – Ana Uspekova Jul 09 '15 at 11:54
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