$AB-I$ is a projection means $BA-I$ is so?

Question

$(AB-I) \text{ is a projection} \Rightarrow (AB-I)^2=AB-I \Rightarrow AB.AB-AB-AB+I=AB-I\\ AB^2A-3AB-3I=0\\ (BA-I)^2= BA^2B-2BA+I$

they are complex matrices. please help

Answer 1

I suppose that both $A$ and $B$ are square matrices. Let $P=AB-I$. Since $Px=-x$ on $\ker B$ but the only possible eigenvalues of a projection are zeroes and ones, we must have $\ker B=0$, i.e. $B$ is invertible. Hence $BA-I=B(AB-I)B^{-1}$ is also a projection.

We have used $-1\ne1$ in the above. The proof actually works over any field of characteristic $\ne2$. When the characteristic is $2$, the statement in question no longer holds. E.g. over $GF(2)$, $$ \pmatrix{1&0\\ 0&0}\pmatrix{0&0\\ 1&0}-I=I $$ is a projection, but $$ \pmatrix{0&0\\ 1&0}\pmatrix{1&0\\ 0&0}-I=\pmatrix{1&0\\ 1&1} $$ is not.

Answer 2

This is true assuming $A,B$ are square matrices.

We have $(AB-I)^2=AB-I$, so $ABAB-3AB=-2I,A(BAB-3B)=-2I$, so $-\frac{1}{2}(BAB-3B)$ is the inverse of $A$. Hence $(BAB-3B)A=-2I,BABA-3BA=-2I$ and $(BA-I)^2=BA-I$.

If we allow non-square matrices, then we can find a counterexample by using $A = (0\, 1)$ and $B=A^T$. Then $AB-I=(0)$ clearly is a projection, but $BA-I$ isn't.