Unstabilizable and undetectable unstable modes break the CARE conclusion

First take $A=C=R=1$ and $B=0$. Then $Q=1$; every feedback leaves $A-BK=1$, so the pair is not stabilizable. The CARE is $2X+1=0$, whose only root is $X=-1\mathrm{/}2$; it is not positive semidefinite and its closed loop remains $1$.

Next take $A=B=R=1$ and $C=0$, so $Q=0$. The pair is stabilizable, since $K=2$ gives $A-BK=-1$, but it is not detectable because $A-LC=1$ for every $L$. The CARE is $2X-X^2=0$, with positive-semidefinite roots $X=0$ and $X=2$; the first has unstable closed loop $1$ and the second stable closed loop $-1$. Since $Q=0$, zero control has zero cost for every initial state, so the value is zero rather than the stabilizing root $X=2$. Thus without detectability uniqueness among PSD roots and the stabilizing value conclusion in [F3] fail.