# Operator norm

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

## Introduction and definition

Given two normed vector spaces ${\displaystyle V}$ and ${\displaystyle W}$ (over the same base field, either the real numbers ${\displaystyle \mathbb {R} }$ or the complex numbers ${\displaystyle \mathbb {C} }$), a linear map ${\displaystyle A:V\to W}$ is continuous if and only if there exists a real number ${\displaystyle c}$ such that [1]

${\displaystyle \|Av\|\leq c\|v\|\quad {\mbox{ for all }}v\in V.}$

The norm on the left is the one in ${\displaystyle W}$ and the norm on the right is the one in ${\displaystyle V}$. Intuitively, the continuous operator ${\displaystyle A}$ never increases the length of any vector by more than a factor of ${\displaystyle c.}$ 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 ${\displaystyle A,}$ it then seems natural to take the infimum of the numbers ${\displaystyle c}$ such that the above inequality holds for all ${\displaystyle v\in V.}$ This number represents the maximum scalar factor by which ${\displaystyle \mathbb {R} ^{n}}$ "lengthens" vectors. In other words, we measure the "size" of ${\displaystyle A}$ by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of ${\displaystyle A}$ as

${\displaystyle \|A\|_{op}=\inf\{c\geq 0:\|Av\|\leq c\|v\|{\mbox{ for all }}v\in V\}.}$

The infimum is attained as the set of all such ${\displaystyle c}$ is closed, nonempty, and bounded from below. [2]

It is important to bear in mind that this operator norm depends on the choice of norms for the normed vector spaces ${\displaystyle V}$ and W.

## Examples

Every real ${\displaystyle m}$-by-${\displaystyle n}$ matrix corresponds to a linear map from ${\displaystyle \mathbb {R} ^{n}}$ to ${\displaystyle \mathbb {R} ^{m}.}$ Each pair of the plethora of (vector) norms applicable to real vector spaces induces an operator norm for all ${\displaystyle m}$-by-${\displaystyle n}$ matrices of real numbers; these induced norms form a subset of matrix norms.

If we specifically choose the Euclidean norm on both ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle \mathbb {R} ^{m},}$ then the matrix norm given to a matrix ${\displaystyle A}$ is the square root of the largest eigenvalue of the matrix ${\displaystyle A^{*}A}$ (where ${\displaystyle A^{*}}$ denotes the conjugate transpose of ${\displaystyle A}$). [3] This is equivalent to assigning the largest singular value of ${\displaystyle A.}$

Passing to a typical infinite-dimensional example, consider the sequence space ${\displaystyle \ell ^{2},}$ which is an Lp space, defined by

${\displaystyle l^{2}=\left\{\left(a_{n}\right)_{n\geq 1}:\;a_{n}\in \mathbb {C} ,\;\sum _{n}|a_{n}|^{2}<\infty \right\}.}$

This can be viewed as an infinite-dimensional analogue of the Euclidean space ${\displaystyle \mathbb {C} ^{n}.}$ Now consider a bounded sequence ${\displaystyle s_{\bullet }=\left(s_{n}\right)_{n=1}^{\infty }.}$ The sequence ${\displaystyle s_{\bullet }}$ is an element of the space ${\displaystyle \ell ^{\infty },}$ with a norm given by

${\displaystyle \left\|s_{\bullet }\right\|_{\infty }=\sup _{n}\left|s_{n}\right|.}$

Define an operator ${\displaystyle T_{s}}$ by pointwise multiplication:

${\displaystyle \left(a_{n}\right)_{n=1}^{\infty }\;{\stackrel {T_{s}}{\mapsto }}\;\ \left(s_{n}\cdot a_{n}\right)_{n=1}^{\infty }.}$

The operator ${\displaystyle T_{s}}$ is bounded with operator norm

${\displaystyle \left\|T_{s}\right\|_{op}=\left\|s_{\bullet }\right\|_{\infty }.}$

This discussion extends directly to the case where ${\displaystyle \ell ^{2}}$ is replaced by a general ${\displaystyle L^{p}}$ space with ${\displaystyle p>1}$ and ${\displaystyle \ell ^{\infty }}$ replaced by ${\displaystyle L^{\infty }.}$

## Equivalent definitions

Let ${\displaystyle A:V\to W}$ be a linear operator between normed spaces. The first four definitions are always equivalent, and if in addition ${\displaystyle V\neq \{0\}}$ then they are all equivalent:

{\displaystyle {\begin{alignedat}{4}\|A\|_{op}&=\inf &&\{c\geq 0~&&:~\|Av\|\leq c\|v\|~&&~{\mbox{ for all }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|\leq 1~&&~{\mbox{ and }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|<1~&&~{\mbox{ and }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|\in \{0,1\}~&&~{\mbox{ and }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|=1~&&~{\mbox{ and }}~&&v\in V\}\;\;\;{\text{ this equality holds if and only if }}V\neq \{0\}\\&=\sup &&{\bigg \{}{\frac {\|Av\|}{\|v\|}}~&&:~v\neq 0~&&~{\mbox{ and }}~&&v\in V{\bigg \}}\;\;\;{\text{ this equality holds if and only if }}V\neq \{0\}.\\\end{alignedat}}}

If ${\displaystyle V=\{0\}}$ then the sets in the last two rows will be empty, and consequently their supremums over the set ${\displaystyle [-\infty ,\infty ]}$ will equal ${\displaystyle \infty }$ instead of the correct value of ${\displaystyle 0.}$ If the supremum is taken over the set ${\displaystyle [0,\infty ]}$ instead, then the supremum of the empty set is ${\displaystyle 0}$ and the formulas hold for any ${\displaystyle V.}$ If ${\displaystyle A:V\to W}$ is bounded then [4]

${\displaystyle \|A\|_{op}=\sup \left\{\left|w^{*}(Av)\right|:\|v\|\leq 1,\left\|w^{*}\right\|\leq 1{\text{ where }}v\in V,w^{*}\in W^{*}\right\}}$

and [4]

${\displaystyle \|A\|_{op}=\left\|{}^{t}A\right\|_{op}}$

where ${\displaystyle {}^{t}A:W^{*}\to V^{*}}$ is the transpose of ${\displaystyle A:V\to W,}$ which is the linear operator defined by ${\displaystyle w^{*}\,\mapsto \,w^{*}\circ A.}$

## Properties

The operator norm is indeed a norm on the space of all bounded operators between ${\displaystyle V}$ and W. This means

${\displaystyle \|A\|_{op}\geq 0{\mbox{ and }}\|A\|_{op}=0{\mbox{ if and only if }}A=0,}$
${\displaystyle \|aA\|_{op}=|a|\|A\|_{op}{\mbox{ for every scalar }}a,}$
${\displaystyle \|A+B\|_{op}\leq \|A\|_{op}+\|B\|_{op}.}$

The following inequality is an immediate consequence of the definition:

${\displaystyle \|Av\|\leq \|A\|_{op}\|v\|\ {\mbox{ for every }}\ v\in V.}$

The operator norm is also compatible with the composition, or multiplication, of operators: if ${\displaystyle V}$, ${\displaystyle W}$ and ${\displaystyle X}$ are three normed spaces over the same base field, and ${\displaystyle A:V\to W}$ and ${\displaystyle B:W\to X}$ are two bounded operators, then it is a sub-multiplicative norm, that is:

${\displaystyle \|BA\|_{op}\leq \|B\|_{op}\|A\|_{op}.}$

For bounded operators on ${\displaystyle V}$, 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

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 ${\displaystyle N^{2}}$ operations (for an ${\displaystyle N\times N}$ matrix), with the exception of the ${\displaystyle \ell _{2}-\ell _{2}}$ norm (which requires ${\displaystyle N^{3}}$ operations for the exact answer, or fewer if you approximate it with the power method or Lanczos iterations).

Computability of Operator Norms [5]
Co-domain
${\displaystyle \ell _{1}}$${\displaystyle \ell _{2}}$${\displaystyle \ell _{\infty }}$
Domain${\displaystyle \ell _{1}}$Maximum ${\displaystyle \ell _{1}}$ norm of a columnMaximum ${\displaystyle \ell _{2}}$ norm of a columnMaximum ${\displaystyle \ell _{\infty }}$ norm of a column
${\displaystyle \ell _{2}}$NP-hardMaximum singular valueMaximum ${\displaystyle \ell _{2}}$ norm of a row
${\displaystyle \ell _{\infty }}$NP-hardNP-hardMaximum ${\displaystyle \ell _{1}}$ norm of a row

The norm of the adjoint or transpose can be computed as follows. We have that for any ${\displaystyle p,q,}$ then ${\displaystyle \|A\|_{p\rightarrow q}=\|A^{*}\|_{q'\rightarrow p'}}$ where ${\displaystyle p',q'}$ are Hölder conjugate to ${\displaystyle p,q,}$ that is, ${\displaystyle 1/p+1/p'=1}$ and ${\displaystyle 1/q+1/q'=1.}$

## Operators on a Hilbert space

Suppose ${\displaystyle H}$ is a real or complex Hilbert space. If ${\displaystyle A:H\to H}$ is a bounded linear operator, then we have

${\displaystyle \|A\|_{op}=\left\|A^{*}\right\|_{op}}$

and

${\displaystyle \left\|A^{*}A\right\|_{op}=\|A\|_{op}^{2},}$

where ${\displaystyle A^{*}}$ denotes the adjoint operator of ${\displaystyle A}$ (which in Euclidean spaces with the standard inner product corresponds to the conjugate transpose of the matrix ${\displaystyle A}$).

In general, the spectral radius of ${\displaystyle A}$ is bounded above by the operator norm of ${\displaystyle A}$:

${\displaystyle \rho (A)\leq \|A\|_{op}.}$

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

However, when a matrix ${\displaystyle N}$ 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

${\displaystyle \rho (N)=\|N\|_{op}.}$

This formula can sometimes be used to compute the operator norm of a given bounded operator ${\displaystyle A}$: define the Hermitian operator ${\displaystyle B=A^{*}A,}$ determine its spectral radius, and take the square root to obtain the operator norm of ${\displaystyle A.}$

The space of bounded operators on ${\displaystyle H,}$ with the topology induced by operator norm, is not separable. For example, consider the Lp space ${\displaystyle L^{2}[0,1],}$ which is a Hilbert space. For ${\displaystyle 0 let ${\displaystyle \Omega _{t}}$ be the characteristic function of ${\displaystyle [0,t],}$ and ${\displaystyle P_{t}}$ be the multiplication operator given by ${\displaystyle \Omega _{t},}$ that is,

${\displaystyle P_{t}(f)=f\cdot \Omega _{t}.}$

Then each ${\displaystyle P_{t}}$ is a bounded operator with operator norm 1 and

${\displaystyle \left\|P_{t}-P_{s}\right\|_{op}=1\quad {\mbox{ for all }}\quad t\neq s.}$

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

## Notes

1. Kreyszig, Erwin (1978), Introductory functional analysis with applications, John Wiley & Sons, p. 97, ISBN   9971-51-381-1
2. See e.g. Lemma 6.2 of Aliprantis & Border (2007).
3. Weisstein, Eric W. "Operator Norm". mathworld.wolfram.com. Retrieved 2020-03-14.
4. Rudin 1991, pp. 92-115.
5. section 4.3.1, Joel Tropp's PhD thesis,

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