Using eigenvectors and values to get systems of differential equations

Question

Can anyone explain how they got these answers? Questions

In #4, my confusion is this: the prior problems involve the eigenvalues via the diagonalization, so not being able to do that leaves me confused.

In #5... I don't see the trick. There's a prior problem where P determined the coefficients and e^(lambda t) and D determined the lambdas, but this appears to do neither (that doesn't correspond to the answer given).

Answer 1

Hint for question 5

$$P^{-1}AP=J \implies A=PJP^{-1}$$

Ans therefore $$A^n=(PJP^{-1})^n=PJ^nP^{-1}$$ Note that $$J=\pmatrix {3 & 1 \\0 & 3}=3\pmatrix{ 1 & 0 \\0 & 1 }+\pmatrix {0 &1 \\0 & 0}$$ $$J=3I_2+B$$

Note also that B is a nilpotent matrix...$B^2=0_2$

Answer 2

I think there is a mistake in the question, the matrix $A$ is supposed to be $A=\begin{pmatrix}0&1\\-9&6\end{pmatrix}$, then the characteristic polynomial is $t^2-6t+9=(t-3)^2$ so it has a double root at t=3. Then A is similar to J and we can write $P^{-1}AP=J$ as suggested. Now you can calculate the matrix exponential as Isham hinted to also get a different answer than was indicated in the text.

Answer 3

There’s no trick to problem 5. In fact, it works exactly the same way that it does in the case that $A$ is diagonalizable. If you have $A=PJP^{-1}$, then $A^n=PJ^nP^{-1}$, which you can see by writing out the product—all of the $P$’s cancel with a $P^{-1}$ except for the two outer ones. So, if you expand $e^At$ by its power series, you get $$\begin{align} I+tA+{t^2\over2!}A^2+{t^3\over3!}A^3+\cdots &= PIP^{-1}+tPJP^{-1}+{t^2\over2!}PJ^2P^{-1}+{t^3\over3!}PJ^3P^{-1}+\cdots \\ &=P\left(I+tJ+{t^2\over2!}J^2+{t^3\over3!}J^3+\cdots\right)P^{-1} \\ &= Pe^{tJ}P^{-1}.\end{align}$$ (I’m playing a bit fast and loose here: in a formal proof these manipulations would require some justification.)

As for problem 4, I suspect that the course material covered something similar, but you can work through it yourself, too. Split $J$ into $3I+N$. These matrices commute, so the familiar real-number exponent rule $e^{M+N}=e^Me^N$ holds for these matrices. You already know how to find the exponential of a diagonal matrix. For $e^{tN}$, observe that $N^2=0$, so the power series of its exponential stops after only two terms, i.e., $e^{tN}=I+tN$, therefore $$e^{tJ}=e^{3tI}(I+tN)=e^{3t}(I+tN).$$