To prove:- $|AB-I|+|BA-I| =0$ given that $A$ is any $3\times 2$ matrix and $B$ is any $2\times 3$ matrix.

Question

I was just told that for any given matrix if it is $n\times (n-1)$ and another of $(n-1)\times n$ then also the above kind of situation that is $$\det(AB-I)+\det(BA-I)=0,$$ where $I$ is the corresponding identity matrix of the square matrix so formed.

Also in order to prove it I tried to solve it by any theorem that I know or so but was unable.

So, I checked it for any random matrix with random variable and calculated the same thing via a matrix calculator, and shockingly it was correct.

I searched it on internet and was unable to find any explanation. So, I eventually turned to stack exchange.

Any help would be appreciated!

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Mar 01 '21 at 06:47
If you look at the answer given below, you'll also see there's nothing special about $n \times (n-1)$ and $(n-1) \times n$ matrices. The only requirement is that the matrices be $m\times n$ and $n\times m$ where $m$ is odd and $n$ is even. — Brian Moehring, Mar 01 '21 at 07:04

Siong Thye Goh · Accepted Answer · 2021-03-02T03:33:14.170

6

\begin{align} \det(AB-I_3) &= (-1)^3\det(I_3 - AB)\\ &= - \det(I_2 - BA)\\ &= - (-1)^2 \det(BA-I_2) \end{align}

The second equality holds due to Sylvester determinant identity.

In general, if $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times m}$, then we have

$$\det(AB-I_m) = (-1)^{n+m} \det(BA-I_n)= (-1)^{n+m \pmod{2}} \det(BA-I_n)$$

edited Mar 02 '21 at 03:33

answered Mar 01 '21 at 06:56

Siong Thye Goh

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To prove:- $|AB-I|+|BA-I| =0$ given that $A$ is any $3\times 2$ matrix and $B$ is any $2\times 3$ matrix.

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