The norm on the left is the one in and the norm on the right is the one in .
Intuitively, the continuous operator never increases the length of any vector by more than a factor of Thus the image of a bounded set under a continuous operator is also bounded. Because of this property, the continuous linear operators are also known as bounded operators.
In order to "measure the size" of one can take the infimum of the numbers such that the above inequality holds for all
This number represents the maximum scalar factor by which "lengthens" vectors.
In other words, the "size" of is measured by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of as
It is important to bear in mind that this operator norm depends on the choice of norms for the normed vector spaces and W.
Every real -by-matrix corresponds to a linear map from to Each pair of the plethora of (vector) norms applicable to real vector spaces induces an operator norm for all -by- matrices of real numbers; these induced norms form a subset of matrix norms.
Passing to a typical infinite-dimensional example, consider the sequence space which is an Lp space, defined by
This can be viewed as an infinite-dimensional analogue of the Euclidean space
Now consider a bounded sequence The sequence is an element of the space with a norm given by
Define an operator by pointwise multiplication:
The operator is bounded with operator norm
This discussion extends directly to the case where is replaced by a general space with and replaced by
Let be a linear operator between normed spaces. The first four definitions are always equivalent, and if in addition then they are all equivalent:
If then the sets in the last two rows will be empty, and consequently their supremums over the set will equal instead of the correct value of If the supremum is taken over the set instead, then the supremum of the empty set is and the formulas hold for any
If is bounded then
where is the transpose of which is the linear operator defined by
The operator norm is indeed a norm on the space of all bounded operators between and . This means
The following inequality is an immediate consequence of the definition:
The operator norm is also compatible with the composition, or multiplication, of operators: if , and are three normed spaces over the same base field, and and are two bounded operators, then it is a sub-multiplicative norm, that is:
For bounded operators on , 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.
Table of common operator norms
Some common operator norms are easy to calculate, and others are NP-hard.
Except for the NP-hard norms, all these norms can be calculated in operations (for an matrix), with the exception of the norm (which requires operations for the exact answer, or fewer if you approximate it with the power method or Lanczos iterations).
In general, the spectral radius of is bounded above by the operator norm of :
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. The quasinilpotent operators is one class of such examples. A nonzero quasinilpotent operator has spectrum So while
This formula can sometimes be used to compute the operator norm of a given bounded operator : define the Hermitian operator determine its spectral radius, and take the square root to obtain the operator norm of