Finding a Basis for S$^\\perp$ - Mathematics Stack Exchange
Nov 5, 2015 · So I was working through this review question and got stumped. My answer isn't completely orthogonal to matrices in a certain subspace, so it's incorrect. The question is: Let …
Prove $AE \perp BE$ using Euclidean construction in a right …
Dec 14, 2025 · I am working on a geometry problem involving a right-angled triangle and a rotation, and I am looking for a purely synthetic (Euclidean) proof. Brainstorming more different …
If $M$ is a closed subspace of an Hilbert space $H$, then $M
I don't manage to prove that $M^ {\perp\perp}\subset M$ if $M$ closed. I know I have to use completeness since there are counterexamples in non-complete spaces, but I am kind of stuck.
finding a basis for $W^\\perp$ and understanding it.
Jul 9, 2013 · Why is $W^\perp = null (A)$ I dont like learning these kinds fo things, is there a way to understand this? WHY is this the case, why do they specifically let A use $w_1$ and $w_2$ …
What is the meaning of superscript $\\perp$ for a vector space
If A A is a matrix, then the nullspace of A A, i.e. null(A) n u l l (A), is a vector subspace. Then, what is the meaning of superscript inverted T T, for example
A question related to $S^ {\perp}$ and closure of span of $S$
Nov 6, 2020 · 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 …
linear algebra - Proof that $W^ {\perp\perp}=W$ (in a finite ...
Jul 10, 2017 · Let $V$ be a finite dimensional vector space over the field $K$, with a non-degenerate scalar product. Let $W$ be a subspace. Show that $W^{\\perp\\perp}=W$. I have ...
Is $V= W\\oplus W^{\\perp}$ always? - Mathematics Stack Exchange
Dec 6, 2020 · Is V = W ⊕W⊥ V = W ⊕ W ⊥ always? Here, W W is a subspace of the vector space V V. I think this holds if V V is finite-dimensional, but what if it is not? Could someone …
The annihilator $M^\perp$ of a set $M \neq \emptyset$ in an …
Jun 6, 2021 · I'm trying to prove the following: Show that the annihilator $M^\perp$ of a set $M \neq \emptyset$ in an inner product space X is a closed subspace of X. Next is the ...
$M^ {\perp \perp}$ is the smallest closed subspace containing M
Oct 29, 2024 · However, since $ H $ is a Hilbert space, $ H = M^ {\perp} \oplus M^ {\perp \perp} $. This means that $ x $ can only be written in one way as the sum of an element in $ M^ {\perp} …