Given two normed vector spacesV and W, a linear map A : V → W is continuous if and only if there exists a real numberc such that The norm on the left is the one in W and the norm on the right is the one in V. Intuitively, the continuous operator A never increases the length of any vector by more than a factor of c. 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 A, it then seems natural to take the infimum of the numbers c such that the above inequality holds for all v in V. In other words, we measure the "size" of A by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of A as The infimum is attained as the set of all such c is closed, nonempty, and bounded from below. It is important to bear in mind that this operator norm depends on the choice of norms for the normed vector spaces V and W.
Examples
Every real m-by-nmatrix corresponds to a linear map from Rn to Rm. Each pair of the plethora of norms applicable to real vector spaces induces an operator norm for all m-by-n matrices of real numbers; these induced norms form a subset of matrix norms. If we specifically choose the Euclidean norm on both Rn and Rm, then the matrix norm given to a matrix A is the square root of the largest eigenvalue of the matrix A*A. This is equivalent to assigning the largest singular value of A. Passing to a typical infinite-dimensional example, consider the sequence space Lp space| defined by This can be viewed as an infinite-dimensional analogue of the Euclidean spaceCn. Now take a bounded sequence s =. The sequence s is an element of the space l∞, with a norm given by Define an operator Ts by simply multiplication: The operator T s is bounded with operator norm One can extend this discussion directly to the case where l2 is replaced by a general Lp space with p > 1 and l∞ replaced by L∞.
Equivalent definitions
One can show that the following definitions are all equivalent if : In the case the sets in the third and fourth row are empty.
Properties
The operator norm is indeed a norm on the space of all bounded operators between V and W. 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 V, W and X are three normed spaces over the same base field, and and are two bounded operators, then it is a sub-multiplicative norm, i.e.: For bounded operators on V, this implies that operator multiplication is jointly continuous. It follows from the definition that a sequence of operators converge in operator norm means they converge 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 N2 operations, with the exception of the norm.
Suppose H is a real or complex Hilbert space. If A : H → H is a bounded linear operator, then we have and where A* denotes the adjoint operator of A. In general, the spectral radius of A is bounded above by the operator norm of A: 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 operatorA has spectrum. So ρ = 0 while. However, when a matrix N is normal, its Jordan canonical form is diagonal ; this is the spectral theorem. In that case it is easy to see that
This formula can sometimes be used to compute the operator norm of a given bounded operatorA: define the Hermitian operatorB = A*A, determine its spectral radius, and take the square root to obtain the operator norm of A. The space of bounded operators on H, with the topology induced by operator norm, is not separable. For example, consider the Hilbert space Lp space|L2. For 0 < t ≤ 1, let Ωt be the characteristic function of , and Pt be the multiplication operator given by Ωt, i.e. Then each Pt is a bounded operator with operator norm 1 and But is an uncountable set. This implies the space of bounded operators on L2 is not separable, in operator norm. One can compare this with the fact that the sequence space l∞ is not separable. The set of all bounded operators on a Hilbert space, together with the operator norm and the adjoint operation, yields a C*-algebra.