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MIT 18.06 Problem Set 7

Eigenvalues and eigenvectors, diagonalization, complex eigenvalues and orthogonality.

1. MIT 18.06 Problem Set 7 — how this assessment works

This lesson reproduces the Problem Set “MIT 18.06 Problem Set 7” from MIT OpenCourseWare 18.06 Linear Algebra (Spring 2010) (https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/), licensed under CC BY-NC-SA 4.0. Work through 12 problems in order: type your reasoning in the answer box or write it on the handwriting pad, then submit for AI grading.

How the grading works Every problem is graded by an AI mathematics teacher that reads the official solution shipped with the course. It checks your work criterion by criterion, awards partial credit for correct steps, and returns written feedback. Equivalent mathematics and valid alternative methods are accepted — a different route to the same result can still earn full marks.

If your answer is incomplete, the feedback names exactly which steps are missing, so you can correct them and try again.

2. MIT 18.06 PSET7 Problem 1 — Fibonacci determinants of the 1, 1, -1 tridiagonal matrix (4 pts)

FnF_n is the determinant of the 1, 1,-1 tridiagonal matrix of order nn:

F2=∣1−111∣=2F3=∣1−1011−1011∣=3F4=∣1−111−111−111∣≠4.F_2=\begin{vmatrix} 1 &-1 \\ 1 & 1 \end{vmatrix}=2 \qquad F_3=\begin{vmatrix} 1 &-1 & 0 \\ 1 & 1 &-1 \\ 0 & 1 & 1 \end{vmatrix}=3 \qquad F_4=\begin{vmatrix} 1 &-1 & & \\ 1 & 1 &-1 & \\ & 1 & 1 &-1 \\ & & 1 & 1 \end{vmatrix} \neq 4.

Expand in cofactors to show that Fn=Fn−1+Fn−2F_n=F_{n-1}+F_{n-2}. These determinants are Fibonacci numbers 1, 2, 3, 5, 8, 13, ... The sequence usually starts 1, 1, 2, 3 (with two 1's) so our FnF_n is the usual Fn+1F_{n+1}.

3. MIT 18.06 PSET7 Problem 2 — Every second Fibonacci number from the 1, 3, 1 recursion (12 pts)

Cofactors of the 1, 3, 1 matrices in Problem 21 give a recursion Sn=3Sn−1−Sn−2S_n=3S_{n-1}-S_{n-2}. Amazingly that recursion produces every second Fibonacci number. Here is the challenge.

Show that SnS_n is the Fibonacci number F2n+2F_{2n+2} by proving F2n+2=3F2n−F2n−2F_{2n+2}=3F_{2n}-F_{2n-2}. Keep using Fibonacci's rule Fk=Fk−1+Fk−2F_k=F_{k-1}+F_{k-2} starting with k=2n+2k=2n+2.

4. MIT 18.06 PSET7 Problem 3 — Pascal matrix with the last entry reduced by 1 (12 pts)

The symmetric Pascal matrices have determinant 1. If I subtract 1 from the nn, nn entry, why does the determinant become zero? (Use rule 3 or cofactors.)

det⁡[1111123413610141020]=1 (known)det⁡[1111123413610141019]=0 (to explain).\det \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{bmatrix}=1 \text{ (known)} \qquad \det \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & \mathbf{19} \end{bmatrix}=\mathbf{0} \text{ (to explain)}.

5. MIT 18.06 PSET7 Problem 4 — Cofactor matrix C and AC^T = (det A)I (4 pts)

Find the cofactors of AA and multiply ACTAC^{T} to find det⁡A\det A:

A=[114122125]andC=[6−30⋅⋅⋅⋅⋅⋅]and ACT=___.A=\begin{bmatrix} 1 & 1 & 4 \\ 1 & 2 & 2 \\ 1 & 2 & 5 \end{bmatrix} \qquad \text{and} \qquad C=\begin{bmatrix} 6 &-3 & 0 \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{bmatrix} \qquad \text{and } AC^{T}=\_\_\_.

If you change that 4 to 100, why is det⁡A\det A unchanged?

6. MIT 18.06 PSET7 Problem 5 — Jacobian of the spherical coordinate transformation (4 pts)

Spherical coordinates ρ\rho, ϕ\phi, θ\theta satisfy x=ρsin⁡ϕcos⁡θx=\rho \sin \phi \cos \theta and y=ρsin⁡ϕsin⁡θy=\rho \sin \phi \sin \theta and z=ρcos⁡ϕz=\rho \cos \phi. Find the 3 by 3 matrix of partial derivatives: ∂x/∂ρ\partial x/\partial \rho, ∂x/∂ϕ\partial x/\partial \phi, ∂x/∂θ\partial x/\partial \theta in row 1. Simplify its determinant to J=ρ2sin⁡ϕJ=\rho^2 \sin \phi. Then dVdV in spherical coordinates is ρ2sin⁡ϕ dρ dϕ dθ\rho^2 \sin \phi \, d\rho \, d\phi \, d\theta the volume of an infinitesimal "coordinate box".

7. MIT 18.06 PSET7 Problem 6 — A "Jacobi formula": 2 by 2 times 3 by 3 minors (12 pts)

Suppose AA is a 5 by 5 matrix. Its entries in row 1 multiply determinants (cofactors) in rows 2–5 to give the determinant. Can you guess a "Jacobi formula" for det⁡A\det A using 2 by 2 determinants from rows 1–2 times 3 by 3 determinants from rows 3–5? Test your formula on the-1, 2,-1 tridiagonal matrix that has determinant 6.

8. MIT 18.06 PSET7 Problem 7 — Cauchy–Binet formula for det(AB) (12 pts)

The 2 by 2 matrix AB=(2 by 3)(3 by 2)AB=(2 \text{ by } 3)(3 \text{ by } 2) has a "Cauchy–Binet formula" for det⁡AB\det AB:

det⁡AB=sum of (2 by 2 determinants in A) (2 by 2 determinants in B).\det AB=\text{sum of (2 by 2 determinants in } A) \ (2 \text{ by 2 determinants in } B).

(a) Guess which 2 by 2 determinants to use from AA and BB.

(b) Test your formula when the rows of AA are 1, 2, 3 and 1, 4, 7 with B=ATB=A^{T}.

9. MIT 18.06 PSET7 Problem 8 — What eigenvalues 0, 1, 2 determine (4 pts)

A 3 by 3 matrix BB is known to have eigenvalues 0,1,20, 1, 2. This is information is enough to find three of these (give the answers where possible):

  • (a) the rank of BB,
  • (b) the determinant of BTBB^{T}B,
  • (c) the eigenvalues of BTBB^{T}B,
  • (d) the eigenvalues of (B2+I)−1(B^2+I)^{-1}.

10. MIT 18.06 PSET7 Problem 9 — Eigenvalues of three 3 by 3 matrices (review) (4 pts)

(Review) Find the eigenvalues of AA, BB, and CC:

A=[123045006]andB=[001020300]andC=[222222222].A=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{bmatrix} \quad \text{and} \quad C=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}.

11. MIT 18.06 PSET7 Problem 10 — All matrices that diagonalize A and A^{-1} (4 pts)

Describe all matrices SS that diagonalize this matrix AA (find all eigenvectors):

A=[4012],A=\begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix},

Then describe all matrices that diagonalize A−1A^{-1}.

12. MIT 18.06 PSET7 Problem 11 — Diagonalizing the transition matrix A_1 (recommended) (4 pts)

(Recommended) Find Λ\Lambda and SS to diagonalize A1A_1 in Problem 15:

A1=[.6.9.4.1].A_1=\begin{bmatrix} .6 & .9 \\ .4 & .1 \end{bmatrix}.

What is the limit of Λk\Lambda^k as k→∞k\to \infty? What is the limit of SΛkS−1S\Lambda^k S^{-1}? In the columns of the matrix you see the ‾\underline{\hspace{2cm}}.

13. MIT 18.06 PSET7 Problem 12 — Left eigenvectors, the rank-1 sum for A, and a Toeplitz challenge (24 pts)

The transpose of A=SΛS−1A=S\Lambda S^{-1} is AT=(S−1)TΛSTA^{T}=(S^{-1})^{T}\Lambda S^{T}. The eigenvectors in ATy=λyA^{T}y=\lambda y are the columns of that matrix (S−1)T(S^{-1})^{T}. They are often called left eigenvectors. How do you multiply matrices to find this formula for AA?

Sum of rank-1 matricesA=SΛS−1=λ1x1y1T+⋯+λnxnynT.\text{Sum of rank-1 matrices} \qquad A=S \Lambda S^{-1}=\lambda_1 x_1 y_1^{T}+\cdots+\lambda_n x_n y_n^{T}.

Challenge problem: in MATLAB (and in GNU Octave), the command A=toepliz(v) produces a symmetric matrix in which each descending diagonal (from left to right) is constant and the first row is v. For instance, if v=[0 1 0 0 0 1]v=[0 \ 1 \ 0 \ 0 \ 0 \ 1], then toepliz(v) is the matrix with 1s on both sides of the main diagonal and on the far corners, and 0s elsewhere. More generally, let v(n)v(n) be the vector in RnR^n with a 1 in the second and last places and 0s elsewhere, and let A(n)=toepliz(v(n))A(n)=\text{toepliz}(v(n)).

(a) Experiment with n=5,…,12n=5, \ldots, 12 in MATLAB to see the repeating pattern of det⁡A(n)\det A(n).

(b) Expand det⁡A(n)\det A(n) in terms of cofactors of the first row and in terms of cofactors of the first column. Use the known determinant CnC_n of problem 5.2.13 to recover the pattern found in part (a).

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