Operator norm

In mathematics, the operator norm is a means to measure the "size" of certain linear operators. 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 V and W (over the same base field, either the real numbers R or the complex numbers C), a linear map A : V → W is continuous if and only if there exists a real number c such that

\|Av\| \le c \|v\| \quad \mbox{ for all } v\in V

(the norm on the left is the one in W, the norm on the right is the one in V). Intuitively, the continuous operator A never "lengthens" any vector more than by a factor of c. 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 smallest number 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 worst case. So we define the operator norm of A as

\|A\|_{op} = \min\{c : \|Av\| \le c \|v\| \mbox{ for all } v\in V\}

(the minimum exists as the set of all such c is closed, nonempty, and bounded from below).

Examples

Every real m-by-n matrix yields a linear map from Rⁿ to R^m. One can put several different norms on these spaces, as explained in the article on norms. Each such choice of norms gives rise to an operator norm and therefore yields a norm on the space of all m-by-n matrices. Examples can be found in the article on matrix norms.

If we specifically choose the Euclidean norm on both Rⁿ and R^m, then we obtain the matrix norm which to a given a matrix A assigns the square root of the largest eigenvalue of the matrix A^*A (where A^* denotes the conjugate transpose of A).

Equivalent definitions

One can show that the following definitions are all equivalent:

\|A\|_{op} = \inf\{c : \|Av\| \le c\|v\| \mbox{ for all } v\in V\}

= \sup\{\|Av\| : v\in V \mbox{ with }\|v\| \le 1\}

= \sup\{\|Av\| : v\in V \mbox{ with }\|v\| = 1\}

= \sup\left\{\frac{\|Av\|}{\|v\|} : v\in V \mbox{ with }v\ne 0\right\}.

Properties

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

\|A\|_{op} \ge 0 \mbox{ and } \|A\|_{op} = 0 \mbox{ if and only if } A = 0

\|aA\|_{op} = |a| \|A\|_{op} \quad\mbox{ for every scalar } a

\|A + B\|_{op} \le \|A\|_{op} + \|B\|_{op}

Furthermore, we have the following important inequality

\|Av\| \le \|A\|_{op} \|v\| \quad\mbox{ for every } v\in V

The operator norm is also compatible with the composition of operators: if V, W and X are three normed spaces over the same base field, and A : V → W and B: W → X are two bounded operators, then

\|BA\|_{op} \le \|B\|_{op} \|A\|_{op}

Operators on a Hilbert space

Suppose H is a real or complex Hilbert space. If A : H → H is a bounded linear operator, then we have

\|A\|_{op} = \|A^*\|_{op}

and

\|A^*A\|_{op} = \|A\|_{op}^2

where A^* denotes the adjoint operator of A (which in finite dimensional situations 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:

\rho(A) \le \|A\|_{op}.

If we have a Hermitian operator H, then using the spectral theorem we can show that

\rho(H) = \|H\|_{op}.

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

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

Gourt Home::Gourt :: Operator norm

Introduction and definition

Examples

Equivalent definitions

Properties

Operators on a Hilbert space

See also