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I have the following equivalent representations for a hyperplane:

$H:=\{x\in\mathbb{R}^n\vert w^T\cdot x=d\},\quad w\in\mathbb{R}^n,d\in\mathbb{R}$ fixed

and

$H_2=\left\{x_0+\sum_{i=1}^{n-1}\lambda_i v_i\,\middle\vert\,\lambda_i\in\mathbb{R}\right\}\subseteq\mathbb{R}^n$ with $v_i$ linearly independent and $x_0\in\mathbb{R}^n$ fixed

In words the first representation of a hyperplane is described by an fixed angle represented by $d$ and a fixed vector given by $w$ so that every $x$ with the scalar product $w^T\cdot x=d$ has to lay on the same hyperplane. So all points belonging to the hyperplane are in $H$.

The $x_0$ in $H_2$ should be equivalent to the $w$ in $H$ and the sum represents a point in the hyperplane as combination of $n-1$ linearly independent vectors, since the vectors are fixed all points have to be on the same hyperplane.

I can describe this equivalency easily with words but I am not able to show it mathematically. Could you please help me with this task?

baxbear
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    $d$ doesn't fix an angle. If fixes a distance to the origin along $w$. – Arthur Aug 08 '19 at 13:21
  • I thought that the scalar product of two vectors represents the position of those two vectors to each other in a given space - maybe not an angle but something similar - so that you can at least say for the same $d$ the angle of the two vectors to each other is the same – baxbear Aug 08 '19 at 13:24
  • If you have a fixed vector $w$, and a fixed real number $d$, then take any vector $x$ such that $w\cdot x = d$, then the component of $x$ which is parallel to $w$ will be the same no matter which $x$ you choose. The length of this component is what $d$ decides (along with the length of $w$). – Arthur Aug 08 '19 at 13:26
  • "then the component of $x$ which is parallel to $w$" I don't really get how these components parallel to a certain vector would describe a plane. In $\mathbb{R}^3$ it describes a cylinder wall. Since all the vectors in the cylinder wall are parallel to w the length wouldn't really matter but the distance to the central vector of the cylinder would be important which could also be called the radius. I doubt that your description is wrong but I don't really get it yet, also the vector product of two parallel vectors is a scalar multiplied by one of those vectors since they are linear dependent. – baxbear Aug 08 '19 at 14:09

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We need that $w\neq 0$, and that $v_1, v_2, \ldots, v_{n-1}$ are linearly independent and all orthogonal to $w$ (implying among other things that $\{w, v_1, v_2, \ldots, v_{n-1}\}$ is a basis of $\Bbb R^n$). And we need $w\cdot x_0 = d$. Once those assumptions are out of the way, we can start showing that the two descriptions give the same hyperplane (if any of these assumptions fail, then they do not describe the same hyperplane).

Choose any $\lambda_i$ for $i$ from $1$ to $n-1$, and calculate $$ w\cdot \left(x_0 + \sum_{i = 1}^{n-1}\lambda_iv_i\right) $$ and you will see that you get $d$. So any vector in $H_2$ will be in $H$.

As for inclusion the other way, take a vector $x$ such that $w\cdot x = d$. Consider the vector $x-x_0$, and write it out as a linear combination in the basis given by $w$ and the $v_i$: $$ x-x_0 = \lambda_0w + \sum_{i = 1}^{n-1}\lambda_iv_i $$ Taking the dot product with $w$ reveals that $\lambda_0 = 0$ (as $w\cdot (x-x_0) = 0$ and $w\cdot v_i = 0$ and $w\cdot w \neq 0$), so we get $$ x-x_0 = \sum_{i = 1}^{n-1}\lambda_iv_i $$ Add $x_0$ to both sides, and you're done.

Arthur
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  • Thank you very much, but could I ask you to maybe add a description to the $H$ representation for a better understanding of mine (as you can see following the comments to the questions I still struggle understanding it) – baxbear Aug 08 '19 at 14:12
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    @baxbear Maybe you should post that as its own question. Be as clear as you can on what exactly it is you don't understand. I have to leave now, but hopefully someone else can give you insightful answers. – Arthur Aug 08 '19 at 14:26