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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.

Ala Tarighati
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Mr Luscombe Maths
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    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 29 '18 at 12:16

4 Answers4

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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 $BA$.

Ala Tarighati
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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.

J.-E. Pin
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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 $$ A=\pmatrix{ 0 & 1 \cr 1 & 0 \cr},\quad B=\pmatrix{ 0 & 1 \cr -1 & 0 \cr}. $$

Dietrich Burde
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  • Note that the OP did not mention that $A$, $B$ are quadratic (Though I think he wants to assume this). – Babelfish Aug 29 '18 at 12:25
  • @Babelfish Yes, you are right, he did not specify this. He also did not specify the domain for the coefficients. I suspect real numbers, though, and quadratic matrices. – Dietrich Burde Aug 29 '18 at 12:26
  • Thank you both. I should have been far more specific in my questioning. @Babelfish, you are correct in your assumptions. – Mr Luscombe Maths Aug 29 '18 at 12:39
  • So does this leave us to conclude that A^{-1}CA is the set of solutions satisfying BA for all real, quadratic matrices, A, such that A is invertible (though as you have already established, this is already defined)? – Mr Luscombe Maths Aug 29 '18 at 12:43
  • This is the right question for you; how much restrictions do we have on $A$? We need $AB=C$, and we need $A$ invertible. Which matrices can $BA$ be? Is $BA=I$ possible? is $BA=0$ possible? is $BA=diag(1/2,1/2)$ possible? – Dietrich Burde Aug 29 '18 at 12:50
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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.

ArsenBerk
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