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Suppose $A$ is a $n \times m$ matrix and $B$ is a $m \times n$ matrix. Then it is known that $det(I_{n}+AB)=det(I_{m}+BA)$.

Is there an analogous identity of the form $det(P_{1}+AB)=det(P_{2}+BA)$, where $P_{1},P_{2}$ are positive definite? Or something like it?

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  • $\begingroup$ Do you have a reference for "it is known that"? $\endgroup$
    – Igor Rivin
    Jul 5, 2012 at 13:21
  • $\begingroup$ This clearly can't be true as stated, if you take $A = B = I_n$ and $P_1 = \lambda I_n$ and $P_2 = \mu I_n$ for two distinct positive numbers $\lambda, \mu$. I don't know if there's something similar. I assume you are working over the field $\mathbb R$. Perhaps your best bet is to diagonalize $P_1$ and $P_2$ and try the explicit formula for the determinant in terms of permutations...? $\endgroup$
    – Spiro Karigiannis
    Jul 5, 2012 at 13:23
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    $\begingroup$ @Igor. See Exercise 14, Chapter 3 of my book *Matrices, Springer-Verlag GTM216. More generally, one has $$X^m\det(XI_n-AB)=X^n\det(XI_m-BA).$$ $\endgroup$
    – Denis Serre
    Jul 5, 2012 at 13:30

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Given $P$, $\det(P+AB)$ does not depend only on $BA$. For example, take $n=m=2$, $P = \pmatrix{2 & 0\cr 0 & 1\cr}$, $A = \pmatrix{1 & t\cr 0 & 1\cr}$, $B = \pmatrix{0 & 1\cr 1 & -t\cr}$. Then $\det(P + AB) = 1 - t$ depends on $t$, but $BA = \pmatrix{0 & 1\cr 1 & 0\cr}$ doesn't depend on $t$. So any $P_2$ such that $\det(P+AB) = \det(P_2+BA)$ must depend on $t$.

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Is this what you are looking for? $$ \det(P+AB)=\det(P)\det(I+P^{-1}AB)=\det P \det(I+BP^{-1}A)$$

Edit: the chain of equalities once ended with $=\det(P+PBP^{-1}A)$, but this only works when all matrices are square.

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  • $\begingroup$ Well, not quite.. but it's a nice one. $\endgroup$
    – Felix Goldberg
    Jul 5, 2012 at 14:22
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    $\begingroup$ $B$ is not a square matrix, so you can't conjugate it by anything. $\endgroup$
    – Will Sawin
    Jul 5, 2012 at 15:36
  • $\begingroup$ @WillSawin: you are right, the last equality does not hold unless $A$ and $B$ are square. But maybe the second-to-last equality (which is basically the Schur complement formula) is a nice enough expression. $\endgroup$
    – Federico Poloni
    Jul 5, 2012 at 17:41

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