Orthonormal basis/Gram-Schmidt [Easy?]

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In summary, the conversation discusses finding an orthonormal basis for the tangent space at points of the form P = (cosht, 0, t) on the regular surface patch σ(t,θ) = (coshtcosθ, coshtsinθ, t). By differentiating and using Gram-Schmidt, it is found that the basis vectors u1 and u2 are already orthogonal. Normalizing them and plugging in θ=0, it is determined that the orthonormal basis is v1 = (sinht, 0, 1) and v2 = (0, cosht, 0).
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sk1001
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Homework Statement



Consider the surface patch σ(t,θ) = (coshtcosθ, coshtsinθ, t) where t is an element of the set of real numbers and θ is an element from (-pi, pi).

Show that σ defines a regular surface patch and find an orthonormal basis for the tangent space (TpS) at points of the form P = (cosht, 0, t)

The Attempt at a Solution


I have done the regular surface patch part. Now just wondering how I go about the orthonormal basis part.

By differentiation:
u1 = ∂σ/dt = (sinhtcosθ, sinhtsinθ, 1) and
u2 = ∂σ/dθ = (-coshtsinθ, coshtcosθ, 1)

Using Gram-Schmidt, I found that u1 and u2 are already orthogonal since the inner product is 0.

I can then normalize u1 and u2:
v1 = u1/|u1| = (sinhtcosθ, sinhtsinθ, 1)/cosht
v2 = u2/|u2| = (-coshtsinθ, coshtcosθ, 1)/cosht

Now I don't know where to go from here...
I thought maybe I should check for a linear combination such that
(cosht, 0, t) = Av1 + Bv2

and then see if the constants A & B satisfy all 3 equations, but that didn't seem to work...
Any other suggestions?
 
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  • #2
You DEFINITELY don't want to try and solve P=Av1+Bv2 or anything like that. P is a point on the surface and v1 and v2 are in the tangent space. They don't really mix. I think you are basically done, except you've got a typo in the last component of u2 and v2, right? The points P=(cosh(t),0,t) are just the points on the surface where theta=0. You could put theta=0 into v1 and v2.
 
  • #3
thanks for the swift reply!
yeah its a typo, last component of u2/v2 should be 0, just an error in copy & paste!

I'm a bit confused with the second part you mentioned, putting theta = 0 into v1 and v2. How did you come to the conclusion that P are points where theta = 0?

Anyhow, if I do this, I'll get...

v1 = (sinht, 0, 1)
v2 = (0, cosht, 0)

Confused!
 
  • #4
sk1001 said:
thanks for the swift reply!
yeah its a typo, last component of u2/v2 should be 0, just an error in copy & paste!

I'm a bit confused with the second part you mentioned, putting theta = 0 into v1 and v2. How did you come to the conclusion that P are points where theta = 0?

Anyhow, if I do this, I'll get...

v1 = (sinht, 0, 1)
v2 = (0, cosht, 0)

Confused!

If I put theta=0 into sigma(t,theta), I get P=(cosh(t),0,t). P is just a curve on the surface where theta=0. (sinh(t),0,1) and (0,cosh(t),0) are an orthonormal basis along that curve. Of course, you did more. You know an orthonormal basis everywhere.
 
  • #5
ah of course, what am I talking about!

so just to make things crystal clear..
σ(t,θ) = (coshtcosθ, coshtsinθ, t)
σ(t,0) = (cosht, 0, t)

Then I have an orthonormal basis
v1 = u1/|u1| = (sinhtcosθ, sinhtsinθ, 1)/cosht
v2 = u2/|u2| = (-coshtsinθ, coshtcosθ, 1)/cosht

Now I plug θ=0 into my orthonal basis vectors and get..
v1 = (sinht, 0, 1)
v2 = (0, cosht, 0)

The end.
Correct?
 
  • #6
sk1001 said:
ah of course, what am I talking about!

so just to make things crystal clear..
σ(t,θ) = (coshtcosθ, coshtsinθ, t)
σ(t,0) = (cosht, 0, t)

Then I have an orthonormal basis
v1 = u1/|u1| = (sinhtcosθ, sinhtsinθ, 1)/cosht
v2 = u2/|u2| = (-coshtsinθ, coshtcosθ, 1)/cosht

Now I plug θ=0 into my orthonal basis vectors and get..
v1 = (sinht, 0, 1)
v2 = (0, cosht, 0)

The end.
Correct?

That's all I can think of to do.
 
  • #7
I was confused before, read the question a little different.
But your suggestion makes a lot more sense when I re-read the question.

Thanks for the help!
 

1. What is an orthonormal basis?

An orthonormal basis is a set of vectors in a vector space that are both orthogonal (perpendicular) and normalized (unit length). In other words, the vectors are all at right angles to each other and have a magnitude of 1.

2. Why is an orthonormal basis important?

An orthonormal basis is important because it allows us to easily represent any vector in a vector space using a linear combination of the basis vectors. This makes computations and transformations in vector spaces much simpler.

3. What is the Gram-Schmidt process?

The Gram-Schmidt process is a method for creating an orthonormal basis from a set of linearly independent vectors. It involves orthogonalizing the vectors by subtracting their projections onto each other, and then normalizing the resulting vectors.

4. How do you use the Gram-Schmidt process to find an orthonormal basis?

To use the Gram-Schmidt process, you start with a set of linearly independent vectors. Then, you apply the process to each vector in the set, creating a new vector that is orthogonal to all previous vectors. Finally, you normalize each resulting vector to obtain an orthonormal basis.

5. Can you give an example of finding an orthonormal basis using the Gram-Schmidt process?

Sure, let's say we have the vectors u = [1, 0, 1] and v = [2, 1, 2]. To find an orthonormal basis, we first normalize u to get u' = [1/√2, 0, 1/√2]. Then, we subtract the projection of v onto u' from v to get v' = [2/√2, 1, 2/√2] - [1/√2, 0, 1/√2] = [3/√2, 1, 3/√2]. Finally, we normalize v' to get v'' = [3/√14, 2/√14, 3/√14]. Thus, our orthonormal basis is {u', v''}.

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