= Solution
Work over a complex <Hilbert space>; for a real space one first complexifies to discuss a complex spectrum. A <bounded linear operator> $L$ is <self-adjoint> when $L=L^*$, where the <adjoint operator> is defined by
$$
\langle Lh,k\rangle=\langle h,L^*k\rangle\qquad(h,k\in H).
$$
Equivalently, $\langle Lh,k\rangle=\langle h,Lk\rangle$ for every pair of vectors.
The <spectrum of a bounded operator> is
$$
\boxed{\Sigma(L)=\{z\in\mathbb C:L-zI\text{ has no bounded, everywhere-defined inverse}\}.}
$$
Its complement is the <resolvent set>. On a <Banach space>, a bounded bijective operator has a <bounded inverse> by the <bounded inverse theorem>, so failure of invertibility is equivalent to failure of bijectivity. An <eigenvalue> is one possible spectral point, but an injective operator that is not onto can also contribute to the <spectrum>.
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