I was reading a book on Linear Control Systems by Prof. Roger Brockett (1970, Wiley).
At the end of Section 1.1, Prof. Brockett asks:
Suppose $K(t)$ is singular for all $t$. Then is
$$
\int\limits_0^T \ K(t) \, dt \ \ \ \mbox{singular}?
$$
Brockett (1970) gives a hint to consider a matrix defined by an outer product
$$
K(t) = \left[ \begin{array}{c}
\sin t \\
\cos t
\end{array} \right] \ \left[ \begin{array}{cc}
\sin t & \cos t \\
\end{array} \right]
$$
and calculate $\int\limits_0^{2 \pi} \ K(t) dt$.
I calculated and found
$$
K(t) = \left[ \begin{array}{cc}
\sin^2 t & \sin t \cos t\\
\sin t \cos t & \cos^2 t
\end{array} \right]
$$
Obviously, $\mbox{det}[K(t)] = 0$ for all $t \in \mathbf{R}$.
Thus, $K(t)$ is singular for all values of $t$.
Moreover,
$$
I = \int\limits_0^{2 \pi} \ \left[ \begin{array}{cc}
\sin^2 t & \sin t \cos t\\
\sin t \cos t & \cos^2 t
\end{array} \right] \ dt = {1 \over 2}
\int\limits_0^{2 \pi} \ \left[ \begin{array}{cc}
1 – \cos 2 t & \sin 2 t \\
\sin 2 t \cos t & 1 + \cos 2 t
\end{array} \right] \ dt
$$
Integrating, we get
$$
I = {1 \over 2} \left[ \begin{array}{cc}
t – {\sin 2 t \over 2} & – {\cos 2 t \over 2} \\[2mm]
– {\cos 2 t \over 2} & t + {\sin 2 t \over 2}
\end{array} \right]_0^{2 \pi} = \left[ \begin{array}{cc}
\pi & 0 \\
0 & \pi
\end{array} \right]
$$
Clearly, $\mbox{det}(I) = \pi^2 \neq 0$.
Thus, the definite integral of a singular matrix need not be singular.
I hope that the calculations (example) are correct. Any other simple example?
Is there any control theoretic interpretation for this example? (Exercise problem)
Best Answer
Here another example:
$K(t)= \left( {\begin{array}{*{20}c} {e^{2t} } & {te^t } \\ {te^t } & {t^2 } \\ \end{array}} \right) $
Of course it is singular for every $t$ but its integral over $[0,2]$ is
$ I = \left[ {\left( {\begin{array}{*{20}c} {\frac{{e^{2t} }}{2}} & {e^t \left( {t - 1} \right)} \\ {e^t \left( {t - 1} \right)} & {\frac{{t^3 }}{3}} \\ \end{array}} \right)} \right]_0^2 = \frac{{e^4 }}{3} - 2e^2 - \frac{7}{3} $