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However, every finite dimensional normed space is a reflexive Banach space, so Riesz’s lemma does holds for = when the normed space is finite-dimensional, as will now be shown. When the dimension of X {\displaystyle X} is finite then the closed unit ball B ⊆ X {\displaystyle B\subseteq X} is compact.
The Gram-Schmidt theorem, together with the axiom of choice, guarantees that every vector space admits an orthonormal basis. This is possibly the most significant use of orthonormality, as this fact permits operators on inner-product spaces to be discussed in terms of their action on the space's orthonormal basis vectors. What results is a deep ...
The norm on induced by , is equal to the original norm on and the continuous dual space of is the set of all real-valued bounded -linear functionals on (see the article about the polarization identity for additional details about this relationship).
Thus as an irreducible representation of SO(3), H ℓ is isomorphic to the space of traceless symmetric tensors of degree ℓ. More generally, the analogous statements hold in higher dimensions: the space H ℓ of spherical harmonics on the n-sphere is the irreducible representation of SO(n+1) corresponding to the traceless symmetric ℓ-tensors.
In finite-dimensional spaces, the matrix representation (with respect to an orthonormal basis) of an orthogonal transformation is an orthogonal matrix. Its rows are mutually orthogonal vectors with unit norm, so that the rows constitute an orthonormal basis of V. The columns of the matrix form another orthonormal basis of V.
In Euclidean space, two vectors are orthogonal if and only if their dot product is zero, i.e. they make an angle of 90° (radians), or one of the vectors is zero. [4] Hence orthogonality of vectors is an extension of the concept of perpendicular vectors to spaces of any dimension.
In other words, the space of orthonormal bases is like the orthogonal group, but without a choice of base point: given the space of orthonormal bases, there is no natural choice of orthonormal basis, but once one is given one, there is a one-to-one correspondence between bases and the orthogonal group.
When X is a vector space and the two metrics and are those induced by norms ‖ ‖ and ‖ ‖, respectively, then strong equivalence is equivalent to the condition that, for all , ‖ ‖ ‖ ‖ ‖ ‖ For linear operators between normed vector spaces, Lipschitz continuity is equivalent to continuity—an operator satisfying either of these ...