Formally, it is a norm defined on the space of bounded linear operators between two given normed vector spaces.
never increases the length of any vector by more than a factor of
This number represents the maximum scalar factor by which
is measured by how much it "lengthens" vectors in the "biggest" case.
matrix corresponds to a linear map from
Each pair of the plethora of (vector) norms applicable to real vector spaces induces an operator norm for all
is the square root of the largest eigenvalue of the matrix
[3] This is equivalent to assigning the largest singular value of
Passing to a typical infinite-dimensional example, consider the sequence space
This can be viewed as an infinite-dimensional analogue of the Euclidean space
is not, in general, guaranteed to achieve its norm
is reflexive if and only if every bounded linear functional
achieves its norm on the closed unit ball.
[4] It follows, in particular, that every non-reflexive Banach space has some bounded linear functional (a type of bounded linear operator) that does not achieve its norm on the closed unit ball.
are three normed spaces over the same base field, and
are two bounded operators, then it is a sub-multiplicative norm, that is:
, this implies that operator multiplication is jointly continuous.
It follows from the definition that if a sequence of operators converges in operator norm, it converges uniformly on bounded sets.
By choosing different norms for the codomain, used in computing
Some common operator norms are easy to calculate, and others are NP-hard.
operations for the exact answer, or fewer if you approximate it with the power method or Lanczos iterations).
is a real or complex Hilbert space.
(which in Euclidean spaces with the standard inner product corresponds to the conjugate transpose of the matrix
To see why equality may not always hold, consider the Jordan canonical form of a matrix in the finite-dimensional case.
Because there are non-zero entries on the superdiagonal, equality may be violated.
is normal, its Jordan canonical form is diagonal (up to unitary equivalence); this is the spectral theorem.
determine its spectral radius, and take the square root to obtain the operator norm of
with the topology induced by operator norm, is not separable.
This implies the space of bounded operators on
One can compare this with the fact that the sequence space