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 …

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?

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 $w_2$ …

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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 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 …

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

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 ...

Mar 25, 2015 · Would a contradiction showing a vector $\vec {v}\in \text {im } (A^T), \notin (\text {ker }A)^\perp$ even be possible? My intuition says yes but the definitions seem to leave no …

Nov 13, 2020 · 1 Let $W$ be a subspace of $\mathbb {R}^n$. We know that $W^\perp$ is also a vector subspace of $\mathbb {R}^n$. How can I show that $\dim W + \dim W^\perp = n$? I …