Observability Gramian

$$W_o(t)={\int }_0^t{e}^{A^T\tau }{C}^{T}C{e}^{At}dt$$

Proof

This is known from the general solution for a LTI system:

$$y(t)=C{e}^{At}x(0)+C{\int }_0^t{e}^{A(t-\tau )}Bu(\tau )d\tau +Du(t)$$

Where the natural response of the system is:

$$y_{nat}(t)=C{e}^{At}x(0)$$

Then: Which gives us:

$$x(0)=W_o^{-1}(t_1){\int }_0^{t_1}{e}^{A^T\tau }{C}^T{y}_{nat}(\tau )d\tau$$

Which proves that the Gramian needs to be a Nonsingular matrix, aka invertible.