2016年8月28日 · In general, a hyperplane in $\Bbb R^n$ is an $(n-1)$-dimensional subspace of $\Bbb R^n$. So, in the case of $\mathbb R^4$, you may think of a hyperplane as a rotated version of our three-dimensional space $\mathbb R^3$. In $\Bbb R^3$, a hyperplane is a two-dimensional plane, and in $\Bbb R^2$, a hyperplane is a one-dimensional line.

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2019年4月1日 · I'm studying about hyperplanes and their equations and had trouble understanding how exactly the hyperplane equation is to be interpreted. I read this answer and this answer and they are great answers no doubt, but I am still having trouble understanding the equation. The definition of a hyperplane given by Boyd is the set

2019年4月5日 · So in the process of constructing a full definition of "hyperplane", the author explicitly imbeds in the as-of-yet incomplete definition of "hyperplane" a term that is defined in terms of the as-of-yet incomplete definition, which therefore presumes that we already have a full definition of the hyperplane.

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2020年10月18日 · Another way to look at this geometrically is with the Notion of scalar product. A hyperplane is the zero locus of a linear functional, as is the space of all vectors orthogonal to the function seen as a vector. (A linear functional is a raw vector, you transpose that and you obtain the vector orthogonal to the hyperplane)

2014年12月16日 · First of all, I think the way you frame the question is confusing. A hyperplane is n-1 dimensional by definition. You should probably be asking "How to prove that this set- Definition of the set H goes here- is a hyperplane, specifically, how to prove it's n-1 dimensional" With that being said. The proof can be separated in two parts:

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2023年8月12日 · $\begingroup$ So for $\mathbb{R}^n$, you find the normal to the $(n-1)$ dimensonial hyperplane by finding out two vectors on the hyperplane, say $(\mathbf{x}, \mathbf{y}) \in \mathbb{R}^{n-1}$, and find the normals as $\mathbf{n} = \mathbf{x} \times \mathbf{y}$; this is understood, right? In the question, I do not understand the role of the ...

First, if we are working over affine space, then the hyperplane is cut out by a global function, so the divisor is principal; in particular, this line bundle is trivial. (In fact, any vector bundle over affine space is trivial, though this is a hard theorem, the Quillen-Suslin theorem .)