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Is the set $Q$ of all operators that have the property that the image is orthogonal to the kernel, and the kernel isn't the null space, only a subset of the set $T$ of the selfadjoint operators or equal to it ? (would this hold in an infinite dimensional space (modulo some closures of the image or kernel) as well ?)

I know that $Q\subseteq T$, but does equality also hold ?

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    I don't understand: every injective operator has the property that the image is orthogonal to the kernel. Or are you asking whether $\operatorname{Im}A \subset (\ker{A})^\perp$ for all self-adjoint $A$? – t.b. Aug 18 '12 at 11:44
  • Yes, but not every selfadjoint operator is injective; so the fact the for selfadjoint operators we still have that the image is orthogonal to the kernel is a more general statement. I just wanted to know if the converse of this statement is also true. –  Aug 18 '12 at 12:35
  • Take a block diagonal matrix consisting of one zero block and another block containing invertible non-symmetric matrix. That's in $Q$ but not in $T$. – t.b. Aug 18 '12 at 12:41
  • Identity is self-adjoint but has null kernel, so it is in $T$ but not $Q$. – tomasz Aug 18 '12 at 12:42

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Take any non-symmetric invertible real matrix. Its image is the whole space, its kernel is zero and these are orthogonal. Yet the operator is not self-adjoint for the standard inner product.

[Added] Your edit does not make much difference, just othogonally project onto a proper finite-dimensional subspace and then apply a non-self-adjoint invertible operator of that subspace to itself; now the kernel is nonzero and orthogonal to the image (the space projected onto) but the operator is not self-adjoint. In spite of your "knowing" it, the claim $Q\subseteq T$ is just false.

Marc van Leeuwen
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  • Sorry, I meant operators whose kernel isn't just the nullspace. I edited my question to hopefully make it more readable. –  Aug 18 '12 at 12:37