On the other hand, since $A,B$ being nonsingular gives us that $AB$ is nonsingular, then we also have that 2) if $AB$ is singular, then at least one of $A$ and $B$ must be singular as well. Notice, by the way, that we also showed that 1) $AB$ is singular if $B$ is the one assumed singular. And that is we we see a singular matrix warning when we try it. So we do not have the ability to write VDV and get something that makes sense. But now we have that $(AB)a=A(Ba)=Ab=0$, and we conclude (again) that $AB$ is singular. But the point is, we do NOT have the relationship that VV is the identity matrix. Now, use that $B$ is nonsingular to find $a$ such that $Ba=b$. If, on the other hand, $B$ is nonsingular, use that $A$ is singular to find $b\ne 0$ such that $Ab=0$. If $B$ is also singular, then for some $x\ne 0$ we have $Bx=0$, but then $(AB)x=A(Bx)=A0=0$, and we conclude that $AB$ is also singular. That $C$ is nonsingular, on the other hand, gives us in particular that the column space of $C$ has full rank, that is, for any vector $b$ there is a vector $a$ such that $Ca=b$. This near-zero matrix is now singular for some maximum lag number (>5) and thus the test crashes. One approach is this: That a matrix $C$ is singular gives us in particular that its null space is non-trivial, that is, for some vector $x\ne0$ we have $Cx=0$. To do this an estimate of the parameters covariance matrix (which is then near-zero) and its inverse is needed (as you can also see in the line invcov np.linalg.inv(covp) in the traceback).
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