How can we prove that an operator which is idempotent but not self adjoint exists?
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2This question and this question tell you exactly where you should look for such operators. – Arthur Apr 28 '20 at 08:54
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Does this answer your question? Normal, idempotent operator implies self-adjointness. – HSN Apr 28 '20 at 10:45
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Claim: An idempotent operator $T:H \to H$, where $H$ is a finite-dimensional inner product space, is self adjoinf if and only if $TT^* = T^*T$.
A proof outline: Let $T$ be an idempotent operator, that is $T^2 = T$. Then $T$ has a basis of eigenvectors with corresponding eigenvalues of $0$ and $1$, as any vector $x$ can be written as: $$x = Tx + (I-T)x.$$ Then $T$ is self-adjoint if and only if the eigenspaces yielded above are mutually orthogonal, which translates to:
\begin{align*} \langle Tx,(I-T)y\rangle=0,\; \forall x,y &\iff \langle x,T^*(I-T)y\rangle = 0, \; \forall x,y \\ &\iff T^*(I-T)y=0,\; \forall y \\ &\iff T^*(I-T) = 0 \\ &\iff T^* = T^*T \end{align*} where the last condition holds if and only if $T^*=T^*T = (T^*T)^*=TT^*$.
Corollary: Any contraction idempotent mapping describes a self-adjoint operator..
Edit: The linked questions by Arthur also contribute to the answer of your question precisely.
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