Although the class was very good about following the other rules, roughly half the class didn’t
follow the leave-me-space-for-comments rule. This particular assignment was mostly computational
and comparatively easy, so in most cases I had very few comments, and so it didn’t matter too
much–on this assignment–whether you left me space for comments. But in the future, please
make sure to leave me the comment-space I’ve instructed you to leave.

    Geometric interpretation of #5. For every v ∈ R³, the map R_v: R³ → R³ defined by R_v(w) = v × w
is linear. If v = 0 then R_v maps every vector w to 0, of course. If v ≠ 0, then every vector w can be expressed uniquely in the form cv + w_⊥, where c ∈ R and w_⊥ is perpendicular to v. Since
R_v(v)= v × v = 0, the “interesting part” of R_v is what it does to vectors orthogonal to v. The set of these vectors is a two-dimensional subspace of R³, the orthogonal complement V^⊥ of the 1-dimensional subspace {all multiples of v}. For every w∈ V^⊥, R_v rotates w by π/2 within the plane V^⊥, and multiplies the length
by ||v||. The sense of the rotation is counterclockwise as seen from the tip of v; i.e. for every nonzero
w ∈ V^⊥, the ordered triple {w, R_v(w), v} is a right-handed triple of mutually orthogonal vectors. For reasons a little beyond the scope of this course, the linear map R_v is called an infinitesimal rotation.
    For p∈R³ and v_p∈T_pR³, we can analogously define the linear map R_vp: T_pR³ → T_pR³ by
R_vp(w_p) = (v × w)_p. The set of equations in problem 5 says that for all s in the domain of β at which the
Frenet frame {T(s), N(s), B(s)} is defined (those s for which κ(s) > 0), the derivative of each element of the
Frenet frame is given by applying the infinitesimal rotation R_A(s) to that element.

(Note: the problem due 2/12/14 originally had “κ_γ/τ_γ” where you now see “τ_γ/κ_γ“. This problem
always should have had “κ_γ/τ_γ“, since τ(t) could be zero for some or all t∈ I, while κ(t) was
assumed nonzero for all t∈ I.)

1. A cubic polynomial with real coefficients has at least one real root. (Hint: if the variable in the
   polynomial is λ, consider what happens as λ →∞ and as λ → –∞, and use the Intermediate
   Value Theorem.)

2. If λ₃ is a real root of the real, cubic polynomial p(λ), then p(λ)/(λ–λ₃) is a quadratic polynomial
   q(λ) with real coefficients.

3. If a quadratic polynomial q(λ) with real coefficients has no real roots, then its roots are a
   complex-conjugate pair a+ bi, a–bi, where b≠ 0.

4. Conclude from the above that if p(λ) is a cubic polynomial with real coefficients, then

   p(λ) = c(λ–λ₁) (λ–λ₂)(λ–λ₃), where c∈R is nonzero, λ₃ ∈R, and λ₁, λ₂ are either both real or are
   complex conjugates of each other.

5. Apply the preceding the to the characteristic polynomial of a 3×3 real matrix A, i.e. the polynomial
   p_A(λ) = det(A–λ I), to show that p_A(λ) = –(λ–λ₁) (λ–λ₂) (λ–λ₃), where λ₁, and λ₂, and λ₃ are as above.
   Recall that λ₁, λ₂, and λ₃ are the eigenvalues of A. Hence A has at least one real eigenvalue λ₃.

6. Recall that det(A)= λ₁ λ₂ λ₃. Hence if A is invertible, which is the case for all orthogonal matrices,
   then it has no zero eigenvalues, so every real eigenvalue is either positive or negative.

7. If A, as above, has a pair of complex-conjugate eigenvalues a± bi, deduce that det(A) = (a² + b²) λ₃,
   and hence that the sign of det(A) is the same as the sign of λ₃. Deduce that (in this case), if
   det(A) > 0 then λ₃ > 0.

8. Since an orthogonal transformations preserve norms, and since there is at least one eigenvector
   for every real eigenvalue, the only possible real eigenvalues of an orthogonal matrix are ±1.

9. If A is the matrix of an orthogonal transformation of R³ and det(A) > 0, then no matter how many real
   eigenvalues A has, at least one of the eigenvalues must be 1 (and there must be an eigenvector with this eigenvalue).

Hint for the remainder of this problem: show that if e is an eigenvector of an orthogonal transformation C,
then C preserves the space of all vectors perpendicular to e (i.e. if v⊥e, then C(v)⊥e), a two-dimensional
subspace (the orthogonal complement of the span of e). Then apply this fact to a basis {e₁, e₂} of this orthogonal complement.

    If you’ve taken complex analysis, the function f in #6 may look familiar to you; it’s the imaginary part
of  –(x + iy)³. Getting rid of the minus sign has the same effect as rotating the surface by π about the
z-axis; it doesn’t change the shape. Similarly, using the real part of (x + iy)³ instead of the imaginary part has the same effect as rotating the surface by π/2 about the z-axis.

“Minimal surfaces” get their name from the following: Let C be a simple closed Curve in R³. Consider
surfaces M in R³ whose boundary is C, where “boundary” here means the set of points in R³ that are not
in M but to which some curve in M gets arbitrarily close. (For example, the equator of a sphere is the
boundary of the open upper hemisphere.) Among all such surfaces, suppose there is one that has smallest
area. Then the mean curvature of this surface is identically 0. (The proof is beyond the scope of this course.)

Problem 5.4/ 3 is an introduction to the subject of geometric partial differential equations. Suppose we ask
the question: find a flat surface, or a minimal surface, subject to some other conditions. (Without other
conditions, a plane would be a cheap answer.) We can start by looking at surfaces that are given as a graph
of a real-valued function f; that’s exactly what a Monge patch gives you. The geometric condition “Gaussian
curvature identically zero” or “mean curvature identically zero” then translates into a nonlinear partial
differential equation for f, which one can try to solve (subject to whatever other conditions are in the problem). Usually, it is extremely difficult to find any closed-form solutions to nonlinear PDEs. Even the
existence/uniqueness theory for solutions of nonlinear PDEs is quite challenging. But often the geometric source of a geometric PDE provides insights that one can use to make clever guesses or simplifications.
The mathematical literature on minimal surfaces alone is vast.

Generalizations of the geometric PDEs in 5.4/3 arise from looking for surfaces of constant (not necessarily
zero) Gaussian curvature or constant mean curvature.