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 ...

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 · However, it doesn't make the matrix become 0 when multiplied, so it's not really a basis for S$^\perp$. Can I get some clarification on what I'm doing wrong, please?

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 answers of different

Jun 21, 2025 · Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges,

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 ...

Dec 6, 2020 · An answer that is easy to digest for an undergraduate who is exposed only to finite-dimensional vector spaces would be nice.

May 14, 2020 · For example, if $U$ is a plane in $\mathbb {R}^3$, then $U^\perp$ is the line normal to that plane. However, it is true in general that $U^\perp$ is a subspace of $V$ whenever $U$ is any …

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 …

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