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Let $A$ be anti-symmetric. That is, $A^T=-A$. Show that the solution to the $\mathbf{x}'=A\mathbf{x}$ satisfies $|\mathbf{x}|=|\mathbf{x}_0|$. Hint: Compute $d/dt|\mathbf{x}|^2$.

The given solution: $d/dt|\mathbf{x}|^2=2\mathbf{x}'\cdot\mathbf{x}=2A\mathbf{x}\cdot\mathbf{x}=2A|\mathbf{x}|^2=0,$ since $A\mathbf{x}\cdot\mathbf{x}=\mathbf{x}A^T\mathbf{x}=-\mathbf{x}A\mathbf{x}.$ Thus, the norm is preserved.

So, I get up until the "since" part. Is that just true because $A$ is anti-symmetric? That man Also, how does that imply that the given quantity is $0?$ I've tried to see it. This is as far as I can go: We have that $A|\mathbf{x}|^2=-\mathbf{x}A\mathbf{x}=-\mathbf{x}\cdot\mathbf{x}'=-e^{tA}\mathbf{x_0}A\mathbf{x}$.

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The expression $2A|\mathbf{x}|^2$ makes no sense as $A$ is a matrix while $|\mathbf{x}|^2$ is a number so let's stop one step before it. The product rule gives us the equation $\frac{d}{dt} |\mathbf{x}|^2 = 2(A\mathbf{x}) \cdot \mathbf{x}$. Assuming that $\mathbf{x}$ is a column vector and calculating the dot product using the definition and $A^T = -A$, we have

$$ (A\mathbf{x}) \cdot \mathbf{x} = (A \mathbf{x})^T \cdot \mathbf{x} = \mathbf{x}^T A^T \mathbf{x} = -\mathbf{x}^T A \mathbf{x} = - \mathbf{x} \cdot (A \mathbf{x}) = -(A \mathbf{x}) \cdot \mathbf{x}$$

which shows that $(A\mathbf{x}) \cdot \mathbf{x} = 0$.

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  • $\begingroup$ You're right, I'm not sure what I was doing... $\endgroup$
    – user269711
    Dec 8, 2016 at 4:42
  • $\begingroup$ I see how that result came about, now. Thank you. $\endgroup$
    – user269711
    Dec 8, 2016 at 4:46

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