9
$\begingroup$

In June 2012, Bill Press and Freeman Dyson published a remarkable paper on the iterated prisoner's dilemma. A key step in their derivation is a simple fact from linear algebra that I feel I should have been explicitly aware of all my life but wasn't: If $M$ is a singular matrix, then we can write down explicit determinantal formulas for some of the vectors in its nullspace. Simply observe that $$M\cdot Adj(M) = (\det M) I = 0$$ where $Adj(M)$ is the adjugate matrix, and observe that the columns of $Adj(M)$ are in the nullspace of $M$.

Now the columns of $Adj(M)$ don't necessarily span the nullspace of $M$ (for a trivial example, take $M=0$). My question is:

Can one write down determinantal formulas for a spanning set of the nullspace of an arbitrary singular matrix $M$? (Perhaps one needs to split into cases and give different formulas in different cases; that's O.K.)

I think that part of the reason I don't already know the answer to this question is that in today's computer age we are obsessed with efficient algorithms, and so we have learned to despise Cramer's rule. However, Press and Dyson's paper illustrates that sometimes this kind of formula can yield important insights.

I should mention that part of the motivation for my question is that Press and Dyson tacitly assume in their paper that for the matrix $M$ they are interested in, the nullspace has dimension one, which strictly speaking is not always true. I think that if the answer to my question is yes, then it should be easy to patch up this gap in their analysis.

$\endgroup$

1 Answer 1

11
$\begingroup$

If the rank of $M$ is $k$ then $\Lambda^kM$ is a ${n\choose k}\times{n\choose k}$ matrix (consisting of al $k\times k$ minors of $M$) of rank $1$. Any of its nonzero rows is a point on $Gr(k,n)$ in its Plucker embedding. This point gives equations of the nullspace.

For example, if $k=1$ then any nonzero row of $M$ gives the equation of the nuspace.

In another example, $k=n-1$, you take $\Lambda^{n-1}M = Adj(M)^T$ and its nonzero rows give you a point of $Gr(n-1,n)$ which is isomorphic to $P^{n-1}$ and under this isomorphism the point gives you the nullspace.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .