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

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

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

• Is that integration defined componentwise? – Daniel Calderón Oct 31 '17 at 5:50

\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".