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I found the eigenvalues and eigenvectors of the matrix $A$, where $$u'(t)=Au$$ is a ODE system with $$A=\left[\begin{array}{ccc} 0 & 3 & 0 \\ -3 & 0 & 4 \\ 0 & -4 & 0 \end{array}\right]$$ But I have to show "without calculus" (as the exercise asks) that the matrix $e^{At}$ is orthogonal and that $||u(t)||^{2}=u_{1}^{2}+u_{2}^{2}+u_{3}^{2}$ is constant.

I really have no idea how can I do it with no calculus.

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    $\begingroup$ I must say it's rather odd to be asked to perform "without calculus" when calculus is used to define the objects of interest--in this case $u' = Au$! Furthermore, $e^{At}$ is usually defined as an infinite series in $At$, so issues of convergence arise, so limits arise, and then you are but a freckle away from calculus! $\endgroup$ – Robert Lewis Jul 18 '17 at 2:06
  • $\begingroup$ Well, the exercise asks just like this: without calculus. $\endgroup$ – mvfs314 Jul 18 '17 at 2:16
  • $\begingroup$ @RobertLewis Hmm, no. What the question asks to prove "without calculus" is not that $t\mapsto e^{tA}$ is a solution of $u'(t)=Au$, but that $e^{tA}$ is orthogonal and $\|e^{tA}u\|^2$ is constant. $\endgroup$ – user1551 Jul 18 '17 at 10:13
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HINT: $(e^{At})^\top = e^{A^\top t} = e^{-At}$.

Next, to show $\|u(t)\|^2$ is constant, you should consider its derivative, namely, $2u(t)\cdot u'(t)$. So what is $Au(t)\cdot u(t)$?

Completely avoiding calculus ...

I suppose you could use write out $e^{At}$ explicitly in terms of those eigenvalues and eigenvectors and just check brute-force. Similarly, you have $u(t)=\sum\limits_{j=1}^3 e^{\lambda_j t}v_j$ (where $v_j$ are the eigenvectors and $\lambda_j$ the corresponding eigenvalues), so you could brute-force compute $\|u(t)\|^2$ and see it's constant.

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    $\begingroup$ It's a standard fact that the solution to $u'=Au$ is given by $u(t)=e^{At}u_0$. Since $e^{At}$ is orthogonal (i.e., norm-preserving) it follows that $\|u(t)\| = \|e^{At}u_0\|=\|u_0\|$, without checking eigenvalues or taking derivatives. I think that this is the whole point of the first part of the exercise. $\endgroup$ – Shalop Jul 18 '17 at 0:58
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    $\begingroup$ That is, since $(e^{At})^T = e^{-At} = (e^{At})^{-1}$, $e^{At}$ is an orthogonal matrix. $\endgroup$ – Robert Israel Jul 18 '17 at 1:11
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    $\begingroup$ Yeah, @Shalop, it's hard to mind-read, but some linear algebra students realize that orthogonal linear maps are isometries. :) $\endgroup$ – Ted Shifrin Jul 18 '17 at 5:03

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