Is every positive semi-definite matrix a controllability Grammian?

Question

Given a stable matrix $A$ (i.e., $A$ has eigenvalues with negative real parts) and a matrix $B$, the (infinite horizon) controllability Grammian $W_{\rm c}(A,B)$ is defined as $$W_{\rm c}(A,B) = \int_0^{\infty} e^{At} B B^T e^{A^T t}$$ Clearly the controllability Grammian is always positive semi-definite. I'm wondering about the converse: is every positive semi-definite matrix a controllability Grammian? Specifically, given a $W \succeq 0$ can we always find $A,B$ such that $W=W_{\rm c}(A,B)$?

Answer 1

For every $W_c$ real, symmetric and positive semidefinite let its Cholesky decomposition $$W_c=L L^T$$ with $L$ a lower triangular matrix with real and nonnegative diagonal entries. Then, selecting $$A=-\rho\mathbb{I}\\ B=\sqrt{2\rho}L$$ with $\rho>0$ we have that $$e^{-\rho\mathbb{I}t}=e^{-\rho t}\mathbb{I}$$ and therefore $$\int_0^{\infty}{e^{At}BB^Te^{A^Tt}dt}=\int_0^{\infty}{e^{-\mathbb{I}\rho t}(\sqrt{2\rho}L)(\sqrt{2\rho}L)^Te^{-\mathbb{I}\rho t}}dt=2\rho\bigg(\int_0^{\infty}{e^{-2\rho t}dt}\bigg) LL^T=W_c$$