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Given a n by n matrix $A$ which is diagonalizable and $a_{ii} = 1$ $\forall i$, I was wondering if the following calculation is possible with trace, \begin{equation} tr \int_0^1 e^{tA} dt = \int_0^1 tr(e^{tA}) dt \end{equation}

We perform an eigenvalue decomposition and get, $tr(Qe^{t\Lambda}Q^{-1})= tr(e^{t\Lambda}) = \sum\limits_{i=0}^n e^{t\lambda_i}$ $\Rightarrow$

$$\int_0^1 \sum\limits_{i=0}^n e^{t\lambda_i} dt= \sum\limits_{i=0}^n\int_0^1 e^{t\lambda_i} dt= \sum\limits_{i=0}^n \frac{e^{\lambda_i} - 1}{\lambda_i} $$ I'm not confidant about the first equality (bringing trace inside integral)

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    $\begingroup$ Can you explain why $tr(e^{tA})=e^{tr(tA)}$? I thought that $det(e^{tA})=e^{tr(tA)}$. $\endgroup$ Aug 14, 2013 at 1:46
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    $\begingroup$ Something is very wrong with your second line. Consider $A = \bigl(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\bigr)$. Then $\operatorname{tr}(e^A) = e+e^{-1}$, whereas $e^{\operatorname{tr}A} = 1$. On the other hand, the trace does happily pass through the integral for finite-dimensional matrices, by linearity of integration. $\endgroup$ Aug 14, 2013 at 2:07
  • $\begingroup$ Thanks @SashaPatotski for pointing that out. I also know that $A$ is diagonalizable, so I added an edit to my question. $\endgroup$
    – user90275
    Aug 14, 2013 at 2:13
  • $\begingroup$ @SashaPatotski, thanks for the edit — I hadn't seen it when I left my comment. Note that my example was diagonal. $\endgroup$ Aug 14, 2013 at 2:17

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As mentioned in the comments, the assertion "$\operatorname{tr}(e^{tA}) = e^{\operatorname{tr}(tA)}$" is simply false.

On the other hand, the integration problem is straightforward. We have $\exp(tA) = \sum_{n\geq 0} A^n \frac{t^n}{n!}$ for any finite-dimensional matrix $A$, since $\exp$ has infinite radius of convergence. By linearity of integration, and checking that certain limits are sufficiently uniform, we have:

$$ \operatorname{tr} \int_0^1 e^{tA}\,\mathrm{d}t = \operatorname{tr} \int_0^1 \sum_{n\geq 0} A^n\, \frac{t^n}{n!}\,\mathrm{d}t = \operatorname{tr}\sum_{n\geq 0} A^n \int_0^1 \frac{t^n}{n!} \,\mathrm d t = \operatorname{tr}\sum_{n\geq 0} \frac{A^n}{(n+1)!} = \operatorname{tr}\left(\frac{e^A-1}{A}\right).$$

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  • $\begingroup$ Thanks Theo! Is yours a more general version of my edit? Is the division by $A$ same as $A^{-1}$? $\endgroup$
    – user90275
    Aug 14, 2013 at 2:20
  • $\begingroup$ @user90275: My formulas are correct for any finite-dimensional matrix, and don't require any diagonalizability. In general, for any analytic function of a single variable $f(x)$, you can define a matrix-valued function $f(A)$ whenever the matrix $A$ has all eigenvalues within the radius of convergence of the Taylor expansion of $f$. An example is $f(x) = \frac{e^x-1}x$, which has infinite radius of convergence. That's what I mean by the formula. $\endgroup$ Aug 14, 2013 at 2:25
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    $\begingroup$ If you'd prefer, I mean $A^{-1}(e^A-1) = (e^A-1)A^{-1}$; note that $A^{-1}$ and $e^A$ commute. But this isn't as good a formula, because $A^{-1}$ may not exist, whereas $\frac{e^A-1}A$ exists for all $A$. $\endgroup$ Aug 14, 2013 at 2:26

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