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 ...

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 ...

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 inner) product …

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 S be a …

Apr 1, 2017 · What is the meaning of superscript $\perp$ for a vector space Ask Question Asked 14 years, 7 months ago Modified 8 years, 8 months ago

Feb 6, 2015 · Showing $ (M^\perp)^\perp=\overline {M}$ Ask Question Asked 10 years, 10 months ago Modified 8 years, 3 months ago

Sep 5, 2019 · As $x \in M^\perp$ was arbitrary, it follows that $\langle m, x \rangle = 0$ holds for all $x \in M^\perp$. (Observe that $m$ satisfies the definition of $M^ {\perp\perp}$).

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.

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 $w_2$ as the …

Nov 22, 2017 · It remains to prove that $ (A^\perp)^\perp $ is the smallest one. Suppose $ F $ is another closed subspace containing $ A $, then $ A^\perp\supset F^\perp $ and hence $ …