Inmathematics, theoperator norm measures the "size" of certainlinear operators by assigning each areal number called itsoperator norm. Formally, it is anorm defined on the space ofbounded linear operators between two givennormed vector spaces. Informally, the operator norm${\displaystyle \Vert T\Vert }$ of a linear map${\displaystyle T:X\to Y}$ is the maximum factor by which it "lengthens" vectors. It is also called thebound norm.^[1]

Given two normed vector spaces${\displaystyle V}$ and${\displaystyle W}$ (over the same basefield, either thereal numbers${\displaystyle \mathbb{R}}$ or thecomplex numbers${\displaystyle \mathbb{C}}$), alinear map${\displaystyle A:V\to W}$ is continuousif and only if there exists a real number${\displaystyle c}$ such that^[2]$${\displaystyle \Vert Av\Vert \le c\Vert v\Vert \text{ for all }v\in V.}$$

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 theimage of a bounded set under a continuous operator is also bounded. Because of this property, the continuous linear operators are also known asbounded operators. In order to "measure the size" of${\displaystyle A,}$ one can take theinfimum 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${\displaystyle A}$ "lengthens" vectors. In other words, the "size" of${\displaystyle A}$ is measured by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of${\displaystyle A}$ as$${\displaystyle \Vert A{\Vert }_{\text{op}}=inf\{c\ge 0:\Vert Av\Vert \le c\Vert v\Vert \text{ for all }v\in V\}.}$$

The infimum is attained as the set of all such${\displaystyle c}$ isclosed,nonempty, andbounded from below.^[3]

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${\displaystyle W}$.

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 ofmatrix norms.

If we specifically choose theEuclidean norm on both${\displaystyle {\mathbb{R}}^n}$ and${\displaystyle {\mathbb{R}}^m,}$ then the matrix norm given to a matrix${\displaystyle A}$ is thesquare root of the largesteigenvalue of the matrix${\displaystyle A^{\ast }A}$ (where${\displaystyle A^{\ast }}$ denotes theconjugate transpose of${\displaystyle A}$).^[4] This is equivalent to assigning the largestsingular value of${\displaystyle A.}$

Passing to a typical infinite-dimensional example, consider thesequence space${\displaystyle {\ell }^2,}$ which is anL^p space, defined by

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

Define an operator${\displaystyle T_s}$ by pointwise multiplication:$${\displaystyle {\left(a_n\right)}_{n=1}^{\mathrm{\infty }}\stackrel{T_s}{\mapsto }{\left(s_n\cdot a_n\right)}_{n=1}^{\mathrm{\infty }}.}$$

The operator${\displaystyle T_s}$ is bounded with operator norm$${\displaystyle {\Vert T_s\Vert }_{\text{op}}={\Vert s_{\bullet }\Vert }_{\mathrm{\infty }}.}$$

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${\displaystyle {\ell }^{\mathrm{\infty }}}$ replaced by${\displaystyle L^{\mathrm{\infty }}.}$

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

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

Importantly, a linear operator${\displaystyle A:V\to W}$ is not, in general, guaranteed to achieve its norm${\displaystyle \Vert A{\Vert }_{\text{op}}=sup\{\Vert Av\Vert :\Vert v\Vert \le 1,v\in V\}}$ on the closed unit ball${\displaystyle \{v\in V:\Vert v\Vert \le 1\},}$ meaning that there might not exist any vector${\displaystyle u\in V}$ of norm${\displaystyle \Vert u\Vert \le 1}$ such that${\displaystyle \Vert A{\Vert }_{\text{op}}=\Vert Au\Vert }$ (if such a vector does exist and if${\displaystyle A\ne 0,}$ then${\displaystyle u}$ would necessarily have unit norm${\displaystyle \Vert u\Vert =1}$). R.C. James provedJames's theorem in 1964, which states that aBanach space${\displaystyle V}$ isreflexive if and only if everybounded linear functional${\displaystyle f\in V^{\ast }}$ achieves itsnorm on the closed unit ball.^[5] 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${\displaystyle A:V\to W}$ is bounded then^[6]$${\displaystyle \Vert A{\Vert }_{\text{op}}=sup\left\{\left|w^{\ast }(Av)\right|:\Vert v\Vert \le 1,\Vert w^{\ast }\Vert \le 1\text{ where }v\in V,w^{\ast }\in W^{\ast }\right\}}$$and^[6]$${\displaystyle \Vert A{\Vert }_{\text{op}}={\Vert {}^{t}A\Vert }_{\text{op}}}$$where${\displaystyle {}^{t}A:W^{\ast }\to V^{\ast }}$ is thetranspose of${\displaystyle A:V\to W,}$ which is the linear operator defined by${\displaystyle w^{\ast }\mapsto w^{\ast }\circ A.}$

The operator norm is indeed a norm on the space of allbounded operators between${\displaystyle V}$ and${\displaystyle W}$. This means$${\displaystyle \Vert A{\Vert }_{\text{op}}\ge 0\text{ and }\Vert A{\Vert }_{\text{op}}=0\text{ if and only if }A=0,}$$$${\displaystyle \Vert aA{\Vert }_{\text{op}}=|a|\Vert A{\Vert }_{\text{op}}\text{ for every scalar }a,}$$$${\displaystyle \Vert A+B{\Vert }_{\text{op}}\le \Vert A{\Vert }_{\text{op}}+\Vert B{\Vert }_{\text{op}}.}$$

The following inequality is an immediate consequence of the definition:$${\displaystyle \Vert Av\Vert \le \Vert A{\Vert }_{\text{op}}\Vert v\Vert \text{ for every }v\in V.}$$

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 asub-multiplicative norm, that is:$${\displaystyle \Vert BA{\Vert }_{\text{op}}\le \Vert B{\Vert }_{\text{op}}\Vert A{\Vert }_{\text{op}}.}$$

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, itconverges uniformly on bounded sets.

By choosing different norms for the codomain, used in computing${\displaystyle \Vert Av\Vert }$, and the domain, used in computing${\displaystyle \Vert v\Vert }$, we obtain different values for the operator norm. Some common operator norms are easy to calculate, and others areNP-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 thepower method orLanczos iterations).

Computability of Operator Norms^[7]

		Co-domain
		${\displaystyle {\ell }_1}$	${\displaystyle {\ell }_2}$	${\displaystyle {\ell }_{\mathrm{\infty }}}$
Domain	${\displaystyle {\ell }_1}$	Maximum${\displaystyle {\ell }_1}$ norm of a column	Maximum${\displaystyle {\ell }_2}$ norm of a column	Maximum${\displaystyle {\ell }_{\mathrm{\infty }}}$ norm of a column
	${\displaystyle {\ell }_2}$	NP-hard	Maximum singular value	Maximum${\displaystyle {\ell }_2}$ norm of a row
	${\displaystyle {\ell }_{\mathrm{\infty }}}$	NP-hard	NP-hard	Maximum${\displaystyle {\ell }_1}$ norm of a row

The norm of theadjoint or transpose can be computed as follows. We have that for any${\displaystyle p,q,}$ then${\displaystyle \Vert A{\Vert }_{p\to q}=\Vert A^{\ast }{\Vert }_{q^{\prime }\to p^{\prime }}}$ where${\displaystyle p^{\prime },q^{\prime }}$ areHölder conjugate to${\displaystyle p,q,}$ that is,${\displaystyle 1/p+1/p^{\prime }=1}$ and${\displaystyle 1/q+1/q^{\prime }=1.}$

Suppose${\displaystyle H}$ is a real or complexHilbert space. If${\displaystyle A:H\to H}$ is a bounded linear operator, then we have$${\displaystyle \Vert A{\Vert }_{\text{op}}={\Vert A^{\ast }\Vert }_{\text{op}}}$$and$${\displaystyle {\Vert A^{\ast }A\Vert }_{\text{op}}=\Vert A{\Vert }_{\text{op}}^2,}$$where${\displaystyle A^{\ast }}$ denotes theadjoint operator of${\displaystyle A}$ (which inEuclidean spaces with the standardinner product corresponds to theconjugate transpose of the matrix${\displaystyle A}$).

In general, thespectral radius of${\displaystyle A}$ is bounded above by the operator norm of${\displaystyle A}$:$${\displaystyle \rho (A)\le \Vert A{\Vert }_{\text{op}}.}$$

To see why equality may not always hold, consider theJordan canonical form of a matrix in the finite-dimensional case. Because there are non-zero entries on the superdiagonal, equality may be violated. Thequasinilpotent operators is one class of such examples. A nonzero quasinilpotent operator${\displaystyle A}$ has spectrum${\displaystyle \{0\}.}$ So${\displaystyle \rho (A)=0}$ while${\displaystyle \Vert A{\Vert }_{\text{op}}>0.}$

However, when a matrix${\displaystyle N}$ isnormal, itsJordan canonical form is diagonal (up to unitary equivalence); this is thespectral theorem. In that case it is easy to see that$${\displaystyle \rho (N)=\Vert N{\Vert }_{\text{op}}.}$$

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

The space of bounded operators on${\displaystyle H,}$ with thetopology induced by operator norm, is notseparable. For example, consider theLp space${\displaystyle L^2[0,1],}$ which is a Hilbert space. For let${\displaystyle {\mathrm{\Omega }}_t}$ be thecharacteristic function of${\displaystyle [0,t],}$ and${\displaystyle P_t}$ be themultiplication operator given by${\displaystyle {\mathrm{\Omega }}_t,}$ that is,$${\displaystyle P_t(f)=f\cdot {\mathrm{\Omega }}_t.}$$

Then each${\displaystyle P_t}$ is a bounded operator with operator norm 1 and$${\displaystyle {\Vert P_t-P_s\Vert }_{\text{op}}=1\text{ for all }t\ne s.}$$

But is anuncountable 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${\displaystyle {\ell }^{\mathrm{\infty }}}$ is not separable.

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

• Banach–Mazur compactum – Concept in functional analysis
• Continuous linear operator – Function between topological vector spaces
• Contraction (operator theory) – Bounded operators with sub-unit norm
• Discontinuous linear map
• Dual norm – Measurement on a normed vector space
• Matrix norm – Norm on a vector space of matrices
• Norm (mathematics) – Length in a vector space
• Normed space – Vector space on which a distance is definedPages displaying short descriptions of redirect targets
• Operator algebra – Branch of functional analysis
• Operator theory – Mathematical field of study
• Topologies on the set of operators on a Hilbert space
• Unbounded operator – Linear operator defined on a dense linear subspace

• Aliprantis, Charalambos D.; Border, Kim C. (2007),Infinite Dimensional Analysis: A Hitchhiker's Guide, Springer, p. 229,ISBN 9783540326960.
• Conway, John B. (1990), "III.2 Linear Operators on Normed Spaces",A Course in Functional Analysis, New York: Springer-Verlag, pp. 67–69,ISBN 0-387-97245-5
• Diestel, Joe (1984).Sequences and series in Banach spaces. New York: Springer-Verlag.ISBN 0-387-90859-5.OCLC 9556781.
• Rudin, Walter (1991).Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY:McGraw-Hill Science/Engineering/Math.ISBN 978-0-07-054236-5.OCLC 21163277.
• Bhatia, Rajendra (1997).Matrix Analysis. Graduate Texts in Mathematics. Vol. 169. New York, NY: Springer New York.doi:10.1007/978-1-4612-0653-8.ISBN 978-1-4612-6857-4.ISSN 0072-5285.

• Functional analysis
• Norms (mathematics)
• Operator theory

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