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Suppose that $\epsilon > 0$, $\gamma: (-\epsilon,\epsilon) \to \text{End}(\mathbb{R}^n)$ is a differentiable curve of matrices such that $\text{det}(\gamma(t)) = 1$ for all $t\in(-\epsilon,\epsilon)$ and $\gamma(0) = I$. Show that $(D\gamma)_0$ has trace zero.

I know that $\gamma(0+h) = \gamma(0)+(D\gamma)_0(h)+\epsilon(h)$ however, I don't know how to calculate the derivative since I don't know the function $\gamma$ explicitly and I am not really sure if there are any properties of the trace that could help me.

McNuggets666
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  • Interesting. This theorem of Liouville should do the trick. Use it to differentiate $\det\gamma(t)$, which you know has derivative zero. – Giuseppe Negro Feb 11 '19 at 16:45
  • @GiuseppeNegro mmm we haven't seen that theorem by Liouville so I don't think that we are able to use it. However, I was thinking that you could use the chain rule in the following way $0=(D1)0 = D(det\circ\gamma)_0 = (Ddet){Id}\circ (D\gamma)_0$,my problem then would be that I think I've heard that the derivative of the determinant is the trace, but I am not sure if this is true. – McNuggets666 Feb 11 '19 at 17:04
  • That's it; the derivative of the determinant is the trace in some sense. Since the derivative of the determinant here is zero, the trace is zero. This is the idea – Giuseppe Negro Feb 11 '19 at 17:05

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Recall that the characteristic polynomial $p_A(t) = \det(tI-A) = t^n -\text{tr}(A)t^{n-1}+\dots+(-1)^n\det A$. From this you should be able to get that $$\frac d{ds}\Big|_{s=0} \det (I+sA) = \text{tr}(A).$$ (Of course, you can get it explicitly by expanding the determinant, too.)

Ted Shifrin
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