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  2. Dot product - Wikipedia

    en.wikipedia.org/wiki/Dot_product

    The name "dot product" is derived from the dot operator " · " that is often used to designate this operation; [1] the alternative name "scalar product" emphasizes that the result is a scalar, rather than a vector (as with the vector product in three-dimensional space).

  3. Vector calculus identities - Wikipedia

    en.wikipedia.org/wiki/Vector_calculus_identities

    For example, from the identity A⋅(B ... The generalization of the dot product formula to Riemannian manifolds is a defining property of a Riemannian connection, ...

  4. Lists of vector identities - Wikipedia

    en.wikipedia.org/wiki/Lists_of_vector_identities

    Vector algebra relations — regarding operations on individual vectors such as dot product, cross product, etc. Vector calculus identities — regarding operations on vector fields such as divergence, gradient, curl, etc.

  5. Vector algebra relations - Wikipedia

    en.wikipedia.org/wiki/Vector_algebra_relations

    The following are important identities in vector algebra.Identities that only involve the magnitude of a vector ‖ ‖ and the dot product (scalar product) of two vectors A·B, apply to vectors in any dimension, while identities that use the cross product (vector product) A×B only apply in three dimensions, since the cross product is only defined there.

  6. Outer product - Wikipedia

    en.wikipedia.org/wiki/Outer_product

    The dot product is the trace of the outer product. [5] ... For example, if is of order 3 ... Another similar identity that further highlights the similarity between ...

  7. Dyadics - Wikipedia

    en.wikipedia.org/wiki/Dyadics

    The dot product takes in two vectors and returns a scalar, ... For example, a dyadic A composed ... sense ii + jj + kk is the identity: ...

  8. Bilinear form - Wikipedia

    en.wikipedia.org/wiki/Bilinear_form

    The dot product on is an example of a bilinear form. [ 1 ] The definition of a bilinear form can be extended to include modules over a ring , with linear maps replaced by module homomorphisms .

  9. Lagrange's identity - Wikipedia

    en.wikipedia.org/wiki/Lagrange's_identity

    Lagrange's identity for complex numbers has been obtained from a straightforward product identity. A derivation for the reals is obviously even more succinct. Since the Cauchy–Schwarz inequality is a particular case of Lagrange's identity, [4] this proof is yet another way to obtain the CS inequality. Higher order terms in the series produce ...