# Intersection of n hyperplanes in $\mathbb{R}^n$

For all unit vector $$\nu \in \mathbb{R}^n$$ consider an affine hyperplane $$A_{\nu}$$ orthogonal to the direction $$\nu$$.

Now consider n linearly independent unit vectors $$\nu_ 1 , \nu_2, \dots, \nu_n \in \mathbb{R}^n$$. I'm asking: is the intersection $$A_{\nu_1} , A_{\nu_2}, \dots , A_{\nu_n}$$ necessarily a point?

I think the answer is yes, but I'm not sure... In the case $$n =2$$ it is true (the intersection of two lines having different directions is a point). In the case $$n=3$$ is also true. But in general is it true?

• When you say that $A_{\nu}$ is an affine hyperplane you must also specify a point on the plane. This does not change the answer to your question. – Eric Towers Dec 5 '18 at 14:24
• Thank you for your contribution Eric. But I don't understand: am I wrong If I say "Consider an affine hyperplane $A_{\nu}$ orthogonal to $\nu$" even if I mean that we don't have information about the point but we only know that it is orthogonal to $\nu$? I hope my question is clear – Hermione Dec 5 '18 at 14:44
• Any plane parallel to the plane you are thinking of is orthogonal to $\nu$. How about the perpendicular bisector of $\nu$? How about the linear subspace perpendicular to $\nu$? – Eric Towers Dec 5 '18 at 16:42

## 1 Answer

Yes, it is just a point. It is a solution of a system$$\left\{\begin{array}{l}\langle\nu_1,x\rangle=\mu_1\\\langle\nu_2,x\rangle=\mu_2\\\vdots\\\langle\nu_n,x\rangle=\mu_n.\end{array}\right.$$This is a system of $$n$$ linear equations in $$n$$ unknowns and the fact that the $$\nu_k$$'s are linearly independent is equivalent to the assertion that the matrix of its coefficients is invertible. Therefore, the system has one and only one solution.

• Is it really $\mu_i$? or 0? – Stockfish Dec 5 '18 at 14:25
• It would be zero if we were working with hyperplanes passing through the origin, but the OP wasn't assuming that. – José Carlos Santos Dec 5 '18 at 14:27
• Ah correct, thank you for the reminder! – Stockfish Dec 5 '18 at 14:28