MIT 18.06 Problem Set 8

This lesson reproduces the Problem Set “MIT 18.06 Problem Set 8” 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 13 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.

The matrix in this question is skew-symmetric ($A^{\mathrm{T}}=-A$):

(a) The derivative of $\|\mathbf{u}(t)\|^2=u_1^2+u_2^2+u_3^3$ is $2u_1u_1'+2u_2u_2'+2u_3u_3'$. Substitute $u_1', u_2', u_3'$ to get zero. Then $\|\mathbf{u}(t)\|^2$ stays equal to $\|\mathbf{u}(0)\|^2$.

(b) When $A$ is skew-symmetric, $Q=e^{At}$ is orthogonal. Prove $Q^{\mathrm{T}}=e^{-At}$ from the series for $Q=e^{At}$. Then $Q^{\mathrm{T}}Q=I$.

Write $A=\begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}$ as $S\Lambda S^{-1}$. Multiply $Se^{\Lambda t}S^{-1}$ to find the matrix exponential $e^{At}$. Check $e^{At}$ and the derivative of $e^{At}$ when $t=0$.

Centering $y''=-y$ in Example 3 will produce $Y_{n+1}-2Y_n+Y_{n-1}=-(\Delta t)^2 Y_n$. This can be written as a one-step difference equation for $\mathbf{U}=(Y, Z)$:

Invert the matrix on the left side to write this as $\mathbf{U}_{n+1}=A\mathbf{U}_n$. Show that $\det A=1$. Choose the large time step $\Delta t=1$ and find the eigenvalues $\lambda_1$ and $\lambda_2=\overline{\lambda_1}$ of $A$:

After 6 steps to $t=6$, $\mathbf{U}_6$ equals $\mathbf{U}_0$. The exact $y=\cos t$ returns to 1 at $t=2\pi$.

The centered choice (leapfrog method) in Problem 28 is very successful for small time steps $\Delta t$. But find the eigenvalues of $A$ for $\Delta t=\sqrt{2}$ and 2:

Both matrices have $|\lambda|=1$. Compute $A^4$ in both cases and find the eigenvectors of $A$. That value $\Delta t=2$ is at the border of instability. Time steps $\Delta t > 2$ will lead to $|\lambda| > 1$, and the powers in $\mathbf{U}_n=A^n \mathbf{U}_0$ will explode.

Note. You might say that nobody would compute with $\Delta t > 2$. But if an atom vibrates with $y''=-1000000y$, then $\Delta t > .0002$ will give instability. Leapfrog has a very strict stability limit. $Y_{n+1}=Y_n+3Z_n$ and $Z_{n+1}=Z_n-3Y_{n+1}$ will explode because $\Delta t=3$ is too large.

Another good idea for $y''=-y$ is the trapezoidal method (half forward/half back): This may be the best way to keep $(Y_n, Z_n)$ exactly on a circle.

(a) Invert the left matrix to write this equation as $\mathbf{U}_{n+1}=A\mathbf{U}_n$. Show that $A$ is an orthogonal matrix: $A^{\mathrm{T}}A=I$. These points $\mathbf{U}_n$ never leave the circle. $A=(I-B)^{-1}(I+B)$ is always an orthogonal matrix if $B^{\mathrm{T}}=-B$.

(b) (Optional MATHLAB) Take 32 steps from $\mathbf{U}_0=(1,0)$ to $\mathbf{U}_{32}$ with $\Delta t=2\pi/32$. Is $\mathbf{U}_{32}=\mathbf{U}_0$? I think there is a small error.

(a) Find a symmetric matrix $\begin{bmatrix} 1 & b \\ b & 1 \end{bmatrix}$ that has a negative eigenvalue.

(b) How do you know it must have a negative pivot?

(c) How do you know it can't have two negative eigenvalues?

Here is a quick "proof" that the eigenvalues of all real matrices are real:

Find the flaw in this reasoning—a hidden assumption that is not justified. You could test those steps on the 90-degree rotation matrix $\begin{bmatrix} 0 &-1 \\ 1 & 0 \end{bmatrix}$ with $\lambda=i$ and $\mathbf{x}=(i, 1)$.

Which of these classes of matrices do $A$ and $B$ belong to: Invertible, orthogonal, projection, permutation, diagonalizable, Markov?

Which of these factorizations are possible for $A$ and $B$: $LU$, $QR$, $S\Lambda S^{-1}$, $Q\Lambda Q^{\mathrm{T}}$?

For complex matrices, the symmetry $A^{\mathrm{T}}=A$ that produces real eigenvalues changes to $\overline{A}^{\mathrm{T}}=A$. From $\det(A-\lambda I)=0$, find the eigenvalues of the 2 by 2 "Hermitian" matrix

To see why eigenvalues are real when $\overline{A}^{\mathrm{T}}=A$, adjust equation (1) of the text to $\overline{A}\overline{\mathbf{x}}=\overline{\lambda}\overline{\mathbf{x}}$.

Transpose to $\overline{\mathbf{x}}^{\mathrm{T}}\overline{A}^{\mathrm{T}}=\overline{\mathbf{x}}^{\mathrm{T}}\overline{\lambda}$. With $\overline{A}^{\mathrm{T}}=A$, reach equation (2): $\lambda=\overline{\lambda}$.

If $\lambda_{max}$ is the largest eigenvalue of a symmetric matrix $A$, no diagonal entry can be larger than $\lambda_{max}$. What is the first entry $a_{11}$ of $A=Q\Lambda Q^T$? Show why $a_{11} \leq \lambda_{max}$.

Prove that the square of a Markov matrix is also a Markov matrix.

A Markov differential equation is not $du/dt=Au$ but $du/dt=(A-I)u$. The diagonal is negative, the rest of $A-I$ is positive. The coulmns add to zero.

Find the eigenvalues of $B=A-I=\begin{bmatrix}-0.2 & 0.3 \\ 0.2 &-0.3 \end{bmatrix}$. Why does $A-I$ have $\lambda=0$?

When $e^{\lambda_1 t}$ and $e^{\lambda_2 t}$ multiply $\mathbf{x}_1$ and $\mathbf{x}_2$, what is the steady state as $t \to \infty$?

(Markov again) This matrix has zero determinant. What are its eigenvalues?

Find the limits of $A^k\mathbf{u}_0$ starting from $\mathbf{u}_0=(1, 0, 0)$ and then $\mathbf{u}_0=(100, 0, 0)$.