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$A$ is an $n$-dimensional square matrix:

Let $X(t)$ be the solution of the problem of initial value $\frac{d}{dt}X=AX$ with $X(t_0)=X_0≠0$. Suppose there exists a real $t_1$ such that $X(t_1)=-X_0$. Prove $A$ has some purely imaginary eigenvalue (non zero).

This was an exam problem from last year in my ODE course and I'm trying to solve it to prepare for my exam.

Carlos
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1 Answers1

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Every square matrix has eigenvalues and we need to demonstrate that one of those eigenvalues of $A$ has a purely imaginary part.

The solution to $\frac{d}{dt}X=AX$ with $X(t_0)=X_0≠0$ is $e^{At}X_0$.

Plug in the second condition, you get $e^{At_1}X_0=-X_0$ which means $-1$ is an eigenvalue of $e^{t_1A}$.

Let $J$ be a Jordan normal form of $A$ and $A=PJP^{-1}$ where $P$ is some invertible square matrix. Let $J_\lambda$ be the block associated with eigenvalue $\lambda$.

Now $e^{At_1}X_0=-X_0$ is equivalent as $Pe^{t_1J}P^{-1}X_0=-X_0$ and therefore $-1$ is an eigenvalue of $e^{t_1J}$.

Jordon canonical blocks have this nice property: the exponentiation of a Jordon block is an upper triangular matrix with the main diagonal equal to the exponentiation of the eigenvalue.

The same property holds for the whole Jordan form.

In turn, the diagonal elements of upper triangular matrices are the eigenvalues of the matrix.

Therefore, one diagonal element of $e^{t_1J}$ is $-1$. Let the Jordon block where $-1$ is located in be in $e^{t_1J}$ be $J_\lambda$. We then have $e^{t_1\lambda}=-1=e^{i(\frac{\pi}{2}+2n\pi)}$ for $n\in\mathbb{Z}$.

Or, $\lambda=\frac{i\pi}{t_1}(\frac{1}{2}+2n)$, which makes $\lambda$ purely imaginary and non-zero.

Argyll
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  • You have shown that there are no real eigenvalues, but how does that assure that there are purely imaginary eigenvalues (real part equal to zero)? – PierreCarre Nov 19 '20 at 18:15
  • @PierreCarre: I can comment on that. – Argyll Nov 19 '20 at 18:19
  • @Argyll So tecnically this could be proven with this same idea? – Carlos Nov 19 '20 at 22:17
  • @Carlos: Yes. Though I am not sure if the linked statement is correct. The argument in the answer doesn't really point to "purely imaginary", just contains non-zero imaginary. Neither does my proof. – Argyll Nov 19 '20 at 22:53
  • @Carlos: I'll add in the part with $-1=e^{i\frac{\pi}{2}+2n\pi}$ for $n\in\mathbb{Z}$ for clarification. – Argyll Nov 19 '20 at 23:01
  • @Argyll Thanks a lot! – Carlos Nov 19 '20 at 23:20
  • @Carlos: Just finished the edit. And you can see answer to both the question with the clarification. It is purely imaginary (or zero) for the other question as well. I was thinking the system can also scale up/down. But any periodicity means purely imaginary eigenvalue. The scaling will mean another eigenvalue has greater than 1 module or something like that. – Argyll Nov 20 '20 at 10:53