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The concept of unit circle (the set of all vectors of norm 1) is different in different norms: for the 1-norm, the unit circle is a square oriented as a diamond; for the 2-norm (Euclidean norm), it is the well-known unit circle; while for the infinity norm, it is an axis-aligned square.
Suppose a vector norm ‖ ‖ on and a vector norm ‖ ‖ on are given. Any matrix A induces a linear operator from to with respect to the standard basis, and one defines the corresponding induced norm or operator norm or subordinate norm on the space of all matrices as follows: ‖ ‖, = {‖ ‖: ‖ ‖ =} = {‖ ‖ ‖ ‖:} . where denotes the supremum.
The Frobenius norm defined by ‖ ‖ = = = | | = = = {,} is self-dual, i.e., its dual norm is ‖ ‖ ′ = ‖ ‖.. The spectral norm, a special case of the induced norm when =, is defined by the maximum singular values of a matrix, that is, ‖ ‖ = (), has the nuclear norm as its dual norm, which is defined by ‖ ‖ ′ = (), for any matrix where () denote the singular values ...
For example, points (2, 0), (2, 1), and (2, 2) lie along the perimeter of a square and belong to the set of vectors whose sup norm is 2. In mathematical analysis , the uniform norm (or sup norm ) assigns, to real- or complex -valued bounded functions f {\displaystyle f} defined on a set S {\displaystyle S} , the non-negative number
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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 .
One way to interpret this theorem is that if the off-diagonal entries of a square matrix over the complex numbers have small norms, the eigenvalues of the matrix cannot be "far from" the diagonal entries of the matrix. Therefore, by reducing the norms of off-diagonal entries one can attempt to approximate the eigenvalues of the matrix.
On a finite-dimensional vector space (but not infinite dimensional vector spaces), all norms are equivalent (although the resulting metric spaces need not be the same) [2] And since any Euclidean space is complete, we can thus conclude that all finite-dimensional normed vector spaces are Banach spaces.