Operator norm
Formally, it is a norm defined on the space of bounded linear operators between two given normed vector spaces.is the maximum factor by which it "lengthens" vectors.is continuous if and only if there exists a real numbernever increases the length of any vector by more than a factor ofThis number represents the maximum scalar factor by whichis measured by how much it "lengthens" vectors in the "biggest" case.matrix corresponds to a linear map fromEach pair of the plethora of (vector) norms applicable to real vector spaces induces an operator norm for allis the square root of the largest eigenvalue of the matrix[3] This is equivalent to assigning the largest singular value ofPassing to a typical infinite-dimensional example, consider the sequence spaceThis can be viewed as an infinite-dimensional analogue of the Euclidean spaceThis discussion extends directly to the case whereis reflexive if and only if every bounded linear functionalachieves 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., 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.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).The norm of the adjoint or transpose can be computed as follows.is a real or complex Hilbert space.(which in Euclidean spaces with the standard inner product corresponds to the conjugate transpose of the matrixTo 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.The quasinilpotent operators is one class of such examples.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 ofwith the topology induced by operator norm, is not separable.One can compare this with the fact that the sequence space