- general topology - The annihilator $M^\perp$ of a set $M \neq \emptyset . . .
I'm trying to prove the following: Show that the annihilator M⊥ M ⊥ of a set M ≠ ∅ M ≠ ∅ in an inner product space X is a closed subspace of X Next is the proof I have done, which indeed only proves that M⊥ M ⊥ can't be open, from what I know, this doesn't imply it is closed, so I need to ensure M⊥ M ⊥ is closed I would really appreciate any corrections to this argument
- 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
- 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 such
- linear algebra - Prove $W \cap W^\perp =\ {\vec {0}\}$ - Mathematics . . .
ok, so how exactly do I take the u (element of) the interesction? I think I know how to take write u is an element of w, and w (perp) sepearately, but no idea what to do with the intersection of them mathematically
- linear algebra - How does one show that a vector in $W$ and $W^ {\perp . . .
"it must be parallel to some vector in W so it can't be perpendicular to W" <- You probably mean it must be a linear combination of some vectors in W?
- Show $(W_1 + W_2 )^\\perp = W_1^\\perp \\cap W_2^\\perp$ and $W_1 . . .
Given V V a inner product space and W1 W 1, W2 W 2 subspaces of V V Show (W1 +W2)⊥ =W⊥1 ∩W⊥2 (W 1 + W 2) ⊥ = W 1 ⊥ ∩ W 2 ⊥ and W⊥1 +W⊥2 ⊆ (W1
- A subspace $X$ is closed iff $X =( X^\\perp)^\\perp$
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- The range of $T^*$ is the orthogonal complement of $\\ker(T)$
How can I prove that, if V V is a finite-dimensional vector space with inner product and T T a linear operator in V V, then the range of T∗ T ∗ is the orthogonal complement of the null space of T T? I know what I must do (for a v v in the range of T∗ T ∗, I have to show that v ⊥ w v ⊥ w for every w w in ker(T) ker (T) and then do the opposite), but I don't know how to show that
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