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Is it true that the trace of a matrix integration equals the integration of the trace of such matrix?

$$\operatorname{tr}\left(\int A(t)\,dt\right) = \int \operatorname{tr}(A(t))\,dt$$

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    $\begingroup$ Is that integration defined componentwise? $\endgroup$ – Daniel Calderón Oct 31 '17 at 5:50
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\begin{align} \operatorname{tr} \int \begin{bmatrix} a_{11}(t) & \cdots & a_{1n}(t) \\ \vdots & & \vdots \\ a_{n1}(t) & \cdots & a_{nn}(t) \end{bmatrix} \, dt & = \operatorname{tr} \begin{bmatrix} \int a_{11}(t)\,dt & \cdots & \int a_{1n}(t) \,dt \\ \vdots & & \vdots \\ \int a_{n1}(t) \, dt & \cdots & \int a_{nn}(t)\,dt \end{bmatrix} \tag 1 \\[10pt] & = \int a_{11}(t)\,dt + \cdots + \int a_{nn}(t)\,dt \tag 2 \\[10pt] & = \int \Big(a_{11}(t) + \cdots + a_{nn}(t)\Big) \, dt \tag 3 \\[10pt] & = \int\operatorname{tr} \begin{bmatrix} a_{11}(t) & \cdots & a_{1n}(t) \\ \vdots & & \vdots \\ a_{n1}(t) & \cdots & a_{nn}(t) \end{bmatrix} \, dt. \tag 4 \end{align} Line $(1)$ makes an assumption about what an integral of a matrix-valued function means.

Line $(2)$ uses the definition of "trace".

Line$(3)$ uses linearity of the integral.

Line $(4)$ uses the definition of "trace".

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If the integration is defined componentwise, it is certainly true -- the integral is the limit of a sum of matrices, and the trace of a sum is the sum of the traces. We can pass to the limit because trace is also continuous.

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