Operator norm explained

In mathematics, the operator norm measures the "size" of certain linear operators by assigning each a real number called its . Formally, it is a norm defined on the space of bounded linear operators between two given normed vector spaces. Informally, the operator norm

\|T\| of a linear map T : X \to Y is the maximum factor by which it "lengthens" vectors.

Introduction and definition

Given two normed vector spaces

V and W (over the same base field, either the real numbers \R or the complex numbers \Complex ), a linear map A : V \to W is continuous if and only if there exists a real number c such that$$\backslash |Av\backslash |\; \backslash leq\; c\; \backslash |v\backslash |\; \backslash quad\; \backslash text\; v\backslash in\; V.$$

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, one can take the infimum of the numbers c such that the above inequality holds for all v \in V. This number represents the maximum scalar factor by which A "lengthens" vectors. In other words, the "size" of A is measured by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of A as$$\backslash |A\backslash |\_\backslash text\; =\; \backslash inf\backslash .$$

The infimum is attained as the set of all such

c is closed, nonempty, and bounded from below.^[1]

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-n matrix corresponds to a linear map from \R ^{n} to \R ^{m.} Each pair of the plethora of (vector) 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

\R ^{n} and \R ^{m,} then the matrix norm given to a matrix A is the square root of the largest eigenvalue of the matrix A ^{*} A (where A ^{*} denotes the conjugate transpose of A ).^[2] This is equivalent to assigning the largest singular value of A. \ell ^{2,} which is an L^p space, defined by$$\backslash ell^2\; =\; \backslash left\backslash .$$\Complex ^{n.} Now consider a bounded sequence s _{\bull} = \left(s _{n\right)} ^{inf ty.} ^{n =1} The sequence s _{\bull} is an element of the space \ell ^{infty,} with a norm given by$$\backslash left\backslash |s\_\backslash right\backslash |\_\backslash infty\; =\; \backslash sup\; \_n\; \backslash left|s\_n\backslash right|.$$

Define an operator

T _{s} by pointwise multiplication:$$\backslash left(a\_n\backslash right)\_^\; \backslash ;\backslash stackrel\backslash ;\backslash \; \backslash left(s\_n\; \backslash cdot\; a\_n\backslash right)\_^.$$

The operator

T _{s} is bounded with operator norm$$\backslash left\backslash |T\_s\backslash right\backslash |\_\backslash text\; =\; \backslash left\backslash |s\_\backslash right\backslash |\_\backslash infty.$$

This discussion extends directly to the case where

\ell ^{2} is replaced by a general L ^{p} space with p > 1 and \ell ^{infty} replaced by L ^{infty.}

Equivalent definitions

Let

A : V \to W be a linear operator between normed spaces. The first four definitions are always equivalent, and if in addition V ≠ \{ 0 \} then they are all equivalent:\begin{alignat}{ 4 } \|A\| _{op} & = inf &&\{ c \geq 0 ~&&:~ \| A v \| \leq c \| v \| ~&&~ forall ~&&v \in V \} \\ & = \sup &&\{ \| Av \| ~&&:~ \| v \| \leq 1 ~&&~ and ~&&v \in V \} \\ & = \sup &&\{ \| Av \| ~&&:~ \| v \| < 1 ~&&~ and ~&&v \in V \} \\ & = \sup &&\{ \| Av \| ~&&:~ \| v \| \in \{ 0 , 1 \} ~&&~ and ~&&v \in V \} \\ & = \sup &&\{ \| Av \| ~&&:~ \| v \| = 1 ~&&~ and ~&&v \in V \} thisequalityholdsifandonlyif V ≠ \{ 0 \} \\ & = \sup && \ { (\| Av \|)/(\| v \|) ~&&:~ v \ne 0 ~&&~ and ~&&v \in V \ } thisequalityholdsifandonlyif V ≠ \{ 0 \}. \\ \end{alignat} If V = \{ 0 \} then the sets in the last two rows will be empty, and consequently their supremums over the set [ -inf ty, inf ty] will equal -inf ty instead of the correct value of 0 . If the supremum is taken over the set [ 0 , inf ty] instead, then the supremum of the empty set is 0 and the formulas hold for any V.

Importantly, a linear operator

A : V \to W is not, in general, guaranteed to achieve its norm \|A\| _{op} = \sup \{\|A v\| : \|v\| \leq 1 , v \in V\} on the closed unit ball \{v \in V : \|v\| \leq 1 \}, meaning that there might not exist any vector u \in V of norm \|u\| \leq 1 such that \|A\| _{op} = \|A u\| (if such a vector does exist and if A ≠ 0 , then u would necessarily have unit norm \|u\| = 1 ). R.C. James proved James's theorem in 1964, which states that a Banach space V is reflexive if and only if every bounded linear functional f \in V ^{*} achieves its norm on the closed unit ball. 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.

If

A : V \to W is bounded then $$\backslash |A\backslash |\_\backslash text\; =\; \backslash sup\; \backslash left\backslash$$and $$\backslash |A\backslash |\_\backslash text\; =\; \backslash left\backslash |^tA\backslash right\backslash |\_\backslash text$$where {} ^{t} A : W ^{*} \to V ^{*} is the transpose of A : V \to W, which is the linear operator defined by w ^{*} \mapsto w ^{*} \circ A.

Properties

The operator norm is indeed a norm on the space of all bounded operators between

V and W . This means$$\backslash |A\backslash |\_\backslash text\; \backslash geq\; 0\; \backslash mbox\; \backslash |A\backslash |\_\backslash text\; =\; 0\; \backslash mbox\; A\; =\; 0,$$$$\backslash |aA\backslash |\_\backslash text\; =\; |a|\; \backslash |A\backslash |\_\backslash text\; \backslash mbox\; a,$$$$\backslash |A\; +\; B\backslash |\_\backslash text\; \backslash leq\; \backslash |A\backslash |\_\backslash text\; +\; \backslash |B\backslash |\_\backslash text.$$

The following inequality is an immediate consequence of the definition:$$\backslash |Av\backslash |\; \backslash leq\; \backslash |A\backslash |\_\backslash text\; \backslash |v\backslash |\; \backslash \; \backslash mbox\backslash \; v\; \backslash in\; V.$$

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 A : V \to W and B : W \to X are two bounded operators, then it is a sub-multiplicative norm, that is: $$\backslash |BA\backslash |\_\backslash text\; \backslash leq\; \backslash |B\backslash |\_\backslash text\; \backslash |A\backslash |\_\backslash text.$$

For bounded operators on

V , 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

By choosing different norms for the codomain, used in computing

\|Av\| , and the domain, used in computing \|v\| , we obtain different values for the operator norm. 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 N ^{2} operations (for an N x N matrix), with the exception of the \ell _{2} - \ell _{2} norm (which requires N ^{3} operations for the exact answer, or fewer if you approximate it with the power method or Lanczos iterations).

Computability of Operator Norms^[3]

		Co-domain
		\ell _{1}	\ell _{2}	\ell _{infty}
Domain	\ell _{1}	Maximum \ell _{1} norm of a column	Maximum \ell _{2} norm of a column	Maximum \ell _{infty} norm of a column
	\ell _{2}	NP-hard	Maximum singular value	Maximum \ell _{2} norm of a row
	\ell _{infty}	NP-hard	NP-hard	Maximum \ell _{1} norm of a row

The norm of the adjoint or transpose can be computed as follows. We have that for any

p, q, then \|A\| _{p →} = \|A ^{*\|} _{q' → p'} where p', q' are Hölder conjugate to p, q, that is, 1 /p + 1 /p' = 1 and 1 /q + 1 /q' = 1 .

Operators on a Hilbert space

Suppose

H is a real or complex Hilbert space. If A : H \to H is a bounded linear operator, then we have$$\backslash |A\backslash |\_\backslash text\; =\; \backslash left\backslash |A^*\backslash right\backslash |\_\backslash text$$and$$\backslash left\backslash |A^*\; A\backslash right\backslash |\_\backslash text\; =\; \backslash |A\backslash |\_\backslash text^2,$$where A ^{*} denotes the adjoint operator of A (which in Euclidean spaces with the standard inner product corresponds to the conjugate transpose of the matrix A ).

In general, the spectral radius of

A is bounded above by the operator norm of A :$$\backslash rho(A)\; \backslash leq\; \backslash |A\backslash |\_\backslash text.$$

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

A has spectrum \{ 0 \}. So \rho(A) = 0 while \|A\| _{op} > 0 .

However, when a matrix

N is normal, its Jordan canonical form is diagonal (up to unitary equivalence); this is the spectral theorem. In that case it is easy to see that$$\backslash rho(N)\; =\; \backslash |N\backslash |\_\backslash text.$$

This formula can sometimes be used to compute the operator norm of a given bounded operator

A : define the Hermitian operator B = 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 Lp space L ^{2[0,} 1 ], which is a Hilbert space. For 0 < t \leq 1 , let \Omega _{t} be the characteristic function of [ 0 , t], and P _{t} be the multiplication operator given by \Omega _{t,} that is,$$P\_t\; (f)\; =\; f\; \backslash cdot\; \backslash Omega\_t.$$

Then each

P _{t} is a bounded operator with operator norm 1 and$$\backslash left\backslash |P\_t\; -\; P\_s\backslash right\backslash |\_\backslash text\; =\; 1\; \backslash quad\; \backslash mbox\; \backslash quad\; t\; \backslash neq\; s.$$

But

\{P _{t} : 0 < t \leq 1 \} is an uncountable set. This implies the space of bounded operators on L ^{2([0,} 1 ]) is not separable, in operator norm. One can compare this with the fact that the sequence space \ell ^{infty} is not separable.

The associative algebra of all bounded operators on a Hilbert space, together with the operator norm and the adjoint operation, yields a C*-algebra.

References

• .
  • Book: Diestel, Joe. Sequences and series in Banach spaces. Springer-Verlag. New York. 1984. 0-387-90859-5. 9556781.

Notes and References

1. See e.g. Lemma 6.2 of .
2. Web site: Operator Norm. Weisstein. Eric W.. Eric W. Weisstein. mathworld.wolfram.com. en. 2020-03-14.
3. section 4.3.1, Joel Tropp's PhD thesis, http://users.cms.caltech.edu/~jtropp/papers/Tro04-Topics-Sparse.pdf