Normals to (hyper)surface must be scalar multiples?

[kingwinner]

Homework Statement

Let S is a (hyper)surface defined by {x|F(x)=0}. Suppose n1 and n2 are both normal to S at x=a. Then n1 and n2 are scalar multiples of each other.

Homework Equations

The Attempt at a Solution

If S is a surface in R³, then I think it's clear geometrically that the above must be true in this case. But what about in higher dimensions (hypersurface)? Is the above still true? If so, how can we prove it? I really can't think of a sound way of justifying it. Or is the above only true for a surface in R³?

Thank you for answering.

 

[susskind_leon]

It basically comes down to the fact that the direction of a vector normal to the surface is given by grad F(x). So yes, it still holds for dim > 3.

 

[kingwinner]

But how do you know that
if there is another normal vector n1 to the (hyper)surface, then it must be a scalar multiple of grad F(x)?

 

[HallsofIvy]

Because there exist only one normal line to a (smooth) surface at any point on that surface. Any normal vector must lie along that line.

 

[kingwinner]

Because there exist only one normal line to a (smooth) surface at any point on that surface. Any normal vector must lie along that line.

I think I understand that for surface, but how about hypersurface? Is there a theorem saying that the noraml line to a smooth hypersurface must be unique?

Thanks.

 

[Dick]

I think I understand that for surface, but how about hypersurface? Is there a theorem saying that the noraml line to a smooth hypersurface must be unique?

Thanks.

A hypersurface in dimension n is given as F(x)=0 where grad(F) is nonzero. Use the implicit function theorem to prove that a hypersurface in dimension n has n-1 independent tangent vectors, if you don't already know that. A normal to the surface has to be orthogonal to all those tangent vectors. That gives you a set of n linearly independent vectors. Can there be two linearly independent normals?