For matrices $A_{m\times n}, B_{n\times m}, m

Question

In manipulations with rectangilar matrices $A_{m\times n}, B_{n\times m}, m<n, m=1,2; n=2,3 $ if we construct square matrices $ P_{m \times m}= AB,~ Q_{n \times n}=BA,$ then it turns out that $$\det(P)=\mbox{Trace}~R~~~~(*)?$$ Here $R=\mbox{adj}~Q,$ or $R$ could also be the the matrix of the cofactors or the minors of $Q$. In this manipulation one can easily explain why $\det(Q)=0$ is always true.

When $m>n$ the roles of $P$ and $Q$ interchange.

The questions are: Can $m$ and $n$ be any natural number? What could be the proof(s) of this (*) interesting result?

ancient mathematician · Accepted Answer · 2019-07-29T07:16:25.010

1

I don't think this is true in general.

Consider, by way of counterexample:

$$ A:=[1,0,0,\dots,0];\ \ B:=A^{T}. $$

Then $P=[1]$ and $\text{adj}\ Q=O$.

Similar counterexamples can be constructed whenever $n>m+1$.

However if $n=m+1$ the result will hold.

If we work with matrices $A,B$ of indeterminates/transcendentals then $P=AB$ can be diagonalised with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_m$.

Also $Q=BA$ will have eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_m,0$ [easy well-known exercise], and (because we are working in the generic case) can also be diagonalised. Then $\text{adj}\ Q$ is diagonal with a single non-zero element $\lambda_1\lambda_2\dots\lambda_m$ and hence we have the result in this generic case. As all the functions in question are polynomial we can substitute for the indeterminates to get the any particular case.

edited Jul 29 '19 at 07:16

answered Jul 29 '19 at 06:59

ancient mathematician

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Thanks, but why adj Q should have diagonal elements as that. – Z Ahmed Jul 29 '19 at 07:09
Let me edit this. – ancient mathematician Jul 29 '19 at 07:13
But my question is not answered still. – Z Ahmed Jul 29 '19 at 07:31
Surely if $Q$ is diagonal, has the eigenvalues I say, then my computation of $\text{adj} Q$ is correct? – ancient mathematician Jul 29 '19 at 08:00
But why should Q be diagonal? – Z Ahmed Jul 29 '19 at 08:09
To compute traces and determinants nothing changes if we diagonalise $Q$, so that is what I say to do. – ancient mathematician Jul 29 '19 at 08:17
Oh! yes you are right. – Z Ahmed Jul 29 '19 at 08:20

For matrices $A_{m\times n}, B_{n\times m}, m

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