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Bookmark or Share- real analysis - A subspace $X$ is closed iff $X =( X^\perp)^\perp ...Which gives X =X⊥⊥ X = X ⊥⊥, as required. (←) (←) If X =X⊥⊥ X = X ⊥⊥ then X X is a closed linear subspace of a Hilbert space. You could try the proof of the converse for yourself. Hint: use a sequence in X X together with the definition of the orthogonal complement, show that the limit point is in X X or X⊥⊥ X ⊥⊥.
- linear algebra - How to explain that null $A$=(row$A$)$^\perp ...That is, x is perpendicular to every row-vector of A. This means in turn that x is in the orthogonal complement to the row space of A. This logic works in reverse as well: if x is in the orthogonal complement of the row-space, then clearly we'll have Ai ⋅ x = 0 for every i, which is to say that Ax = 0. So, null (A) = row (A) ⊥.
- What is the meaning of superscript $\\perp$ for a vector spaceStack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
- $(U\\cap W)^\\perp = U^\\perp + W^\\perp$ in metric vector spacesSide note: the term "metric vector space" is, as far as I know, not in extremely widespread use and might even be peculiar to the Roman textbook.
- real analysis - show that $A^{\perp} = (\overline{A})^{\perp ...This question shows research effort; it is useful and clear. 1. A is a non-empty subset of a pre-Hilbert space. and I'm asked to show that the orthogonal complement of A is equal to the orthogonal complement of its closure. I solved it in a different way than what the solution of the textbook suggests so I'm just asking for confirmation if my ...
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