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  • linear algebra - Proving that $M^ {\perp} = M^ {\perp\perp\perp . . .
    Let M M be a non-empty subset of a Hilbert space H H First, prove that M ⊂M⊥⊥ M ⊂ M ⊥⊥ I know it must be trivial but I still cannot wrap my head around it Why can't we just like that claim M⊥⊥ ⊂ M M ⊥⊥ ⊂ M? Also, is it true for non-complete inner product spaces? Second, prove that M⊥ =M⊥⊥⊥ M ⊥ = M ⊥⊥⊥ This one I have no idea how to approach
  • Prove that Euclidean space is the direct sum of $U$ and $U^\\perp$
    Prove that Euclidean space is the direct sum of U U and U⊥ U ⊥, where U U and U⊥ U ⊥ are the subspaces of E E Because we know that dim E = dim U + dimU⊥ − dim(U ∩U⊥) dim E = dim U + dim U ⊥ − dim (U ∩ U ⊥) is it enough to prove that the intersection of U U and U⊥ U ⊥ is the null-space? That would mean that dim E = dim U + dimU⊥ dim E = dim U + dim U ⊥ and so
  • linear algebra - Show that $ (U + W)^ {\perp} = U^ {\perp}\cap W . . .
    If U U and W W are subspaces of a finite dimensional inner product space V V, show that (a) If U ⊆ W U ⊆ W, then W⊥ ⊆ U⊥ W ⊥ ⊆ U ⊥ (b) (U + W)⊥ =U⊥ ∩W⊥ (U + W) ⊥ = U ⊥ ∩ W ⊥ (c) U⊥ +W⊥ ⊂ (U ∩ W)⊥ U ⊥ + W ⊥ ⊂ (U ∩ W) ⊥ MY ATTEMPT (a) Let {u1,u2, …,um} {u 1, u 2, …, u m} be a basis for U U Since U U is a subspace of W W, we can extend
  • A question related to $S^{\\perp}$ and closure of span of $S$
    This question was asked in my linear algebra quiz previous year exam and I was unable to solve it Let V be an inner ( in question it's written integer , but i think he means inner) product space
  • How to prove $ A^{\\perp} $ is a closed linear subspace?
    Suppose $ X $ is an inner product space and $ A\\subseteq X $ I need to prove that $ A^{\\perp} $ is a closed linear subspace of $ X $ Can anyone give me a idea?
  • Show $(W_1 + W_2 )^\\perp = W_1^\\perp \\cap W_2^\\perp$ and $W_1 . . .
    Given V a inner product space and W1, W2 subspaces of V Show (W1 + W2) ⊥ = W ⊥ 1 ∩ W ⊥ 2 and W ⊥ 1 + W ⊥ 2 ⊆ (W1 ∩ W2) ⊥
  • What is the meaning of superscript $\\perp$ for a vector space
    It A⊥ A ⊥ means orthogonal complement of A A, meaning the subspace that consists of all vectors which when dotted with any vector from A A produce 0 0, that is
  • linear algebra - $ (U\cap W)^\perp = U^\perp + W^\perp$ in metric . . .
    I'm currently working my way through Roman Advanced Linear Algebra chapter 11 and am getting caught up on 2 (b) Let U, W U, W be subspaces of a metric vector space Prove that (U ∩ W)⊥ =U⊥ +W⊥ (U ∩ W) ⊥ = U ⊥ + W ⊥ I've proven that U⊥ +W⊥ ⊆ (U ∩ W)⊥ U ⊥ + W ⊥ ⊆ (U ∩ W) ⊥, but I'm struggling to prove the other way around Every proof I've found so far on
  • real analysis - What is this upside-down T notation: $S^\perp . . .
    In this screenshot, I want to know that the upside-down T is (I'm not sure how to research it if I don't know its name) =) (The context, is to prove that that $ (S^ {\perp})^ {\perp}$ is the smallest





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