show/hide this revision's text 3 added 97 characters in body

Recall that the spectrum of an operator $A$ on a Hilbert space is the set of vales $\lambda$ such that $A-\lambda I$ does not have a bounded inverse . So if $A$ is multiplication by a function in $L^2$ of a measure space, then any point of the essential range of the function is in the spectrum. So, for example (and this is the classic example) the spectrum of multiplication by $x$ on the real line is the whole line. However, a point of the spectrum is not necessarily an eigenvalue. In fact $\lambda$ is an eigenvalue (or in the point spectrum) iff $A- \lambda I$ has a non-trivial null space. (And as others have pointed out, on a separable Hilbert space, the point spectrum is countable.)

show/hide this revision's text 2 added 221 characters in body; added 3 characters in body; added 1 characters in body

Recall that the spectrum of an operator $A$ on a Hilbert space is the set of vales $\lambda$ such that $A-\lambda I$ is does not invertiblehave a bounded inverse . So if $A$ is multiplication by a function in $L^2$ of a measure space, then any point of the essential range of the function is in th the spectrum. So, for example (and this is the classic example) the spectrum of multiplication by $x$ on the real line is the whole line. However, a point of the spectrum is not necessarily an eigenvalue. In fact $\lambda$ is an eigenvalue (or in the point spectrum) iff $A- \lambda I$ has a non-trivial null space.

show/hide this revision's text 1

Recall that the spectrum of $A$ is the set of vales $\lambda$ such that $A-\lambda I$ is not invertible. So if $A$ is multiplication by a function in $L^2$ of a measure space, then any point of the essential range of the function is in th spectrum. So, for example (and this is the classic example) the spectrum of multiplication by $x$ on the real line is the whole line.