:Talk:Dual space/Archive 2

The article currently says that A^0 + B^0 = (A \cap B)^0 and equality holds when V is finite-dimensional. This is always true, and the only reason to assume finite-dimensionality is to make this easier to prove. If we are not providing a proof anyway, then I don't see any reason to assume that V is finite-dimensional. 50.1.105.93 (talk) 06:24, 7 July 2016 (UTC)

"In mathematics, any vector space V has a corresponding dual vector space (or just dual space for short) consisting of all linear functionals on V together with a naturally induced linear structure."

Here, "naturally induced linear structure" is meaningless, so this sentence does not serve as a definition as the following sentence claims. 96.255.35.216 (talk) 13:11, 19 August 2017 (UTC)

In the second paragraph under "Algebraic dual space", it says, "The pairing of a functional $\backslash phi$ in the dual space $V^*$ and an element $x$ of $V$ ... is called the natural pairing." Yet it doesn't explain what this natural pairing is, or under what conditions it exists. The linked article on a dual pair seems to suggest the word "pair" refers to the spaces themselves, not the vectors within. Is there some natural way to match vectors when the vector spaces form a dual pair?

Tnedde (talk) 15:35, 16 June 2018 (UTC)

:Since the bilinear mapping defined by {{nowrap|{{langle}}·,·{{rangle}} : V^∗ × V → F}} maps all ordered pairs from the cross product, V^∗ × V, it is clear that any functional φ in the dual space can be paired with any element x of V. Thus all such pairs are natural pairings.—Anita5192 (talk) 03:38, 17 June 2018 (UTC)

In the very first section after the table of contents, it's suggested that arguments in square brackets be ordered thus [x,φ] and in angled brackets thus <φ,x>. But by the time we get to 'Transpose of a linear map', you are using square brackets with the second order. 37.205.58.146 (talk) 12:58, 1 May 2019 (UTC)