Given the matrix $AB$, find the matrix $BA$

Question

I have that $$AB = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$ and I am asked to find $BA$. We know that $A$ and $B$ are $2\times 2$ matrices with real number entries.

Please could I ask how people might approach this problem?

So far, I have that the solutions are given by $A^{-1}B^{-1}$. We can establish from the initial information that $A$ and $B$ are both invertible. Is there a way of showing that this set of solutions is finite or infinite?

I know that $A$ and $B$ are invertible since $\det(BA)=\det(A)\det(B)$ is non-zero. Credit goes to the poster who provided such an eloquent explanation to get my current level of understanding.

Answer 1

Let $AB=C$ with $C=\pmatrix{ -1 & 0 \cr 0 & 1 \cr}$. Since $0\neq \det(C)=\det(A)\det(B)$, both $A$ and $B$ are invertible. Conjugating $AB=C$ with $A$ yields $$ BA=A^{-1}(AB)A=A^{-1}CA. $$

Note that there are several choices for $A$ and $B$ with $AB=C$. For example, $A=I$, $B=C$, or $B=I$ and $A=C$. Another possibility is $$ A=\pmatrix{ 0 & 1 \cr 1 & 0 \cr},\quad B=\pmatrix{ 0 & 1 \cr -1 & 0 \cr}. $$

Answer 2

HINT: Take $A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$ and $B = \begin{pmatrix} e & f \\ g & h \\ \end{pmatrix}$. Then by using the equation given, find some restrictions for the unknown variables $a,b,c,d,e,f,g,h$. After that, you can find $BA$ in terms of unknown variables and with the restrictions, you can generalize some of the results.

Answer 3

A trivial answer to this question is as follows:

Let us assume $A=I_2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then $B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$.

Which means $AB=BA= \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}.$

I think there are multiple answers to this question and we need more information.

Also it is not mentioned anywhere that matrices $A$ and $B$ should have dimension $2\times 2$, which means we can have $A$ of dimension $m\times 2$ and $B$ of dimension $2\times n$ for $n\neq m$, which means we cannot find $BA$.

Answer 4

From $$AB = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$ follows $ABAB = I$ where $I$ is the identity matrix. Consequently

1. $A$ and $B$ are invertible,
2. $A(BA)B = I$, whence $BA = A^{-1}B^{-1}$.

Thus the answer is $BA = A^{-1}B^{-1}$. It is the unique solution and it does not require any further information.