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Can anyone prove this using only linear combination/matrix multiplication ?

Let $A$ and $B$ be $2 X 2$ matrices such that $AB = I$. Prove that $BA = I.$

$\left[\begin{matrix} a&b\\c&d \end{matrix}\right]\left[\begin{matrix} e&f\\g&h \end{matrix}\right]=\left[\begin{matrix} 1&0\\0&1 \end{matrix}\right]$ is of no use as I end up with so many unknowns

$AB=\left[\begin{matrix} A_{first row}B_{first column}&A_{first row}B_{second column}\\A_{second row}B_{first column}&A_{second row}B_{second column} \end{matrix}\right]$ is also not helping

Vikram
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2 Answers2

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The trick is to first show that if $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $AB = I$ then necessarily $B = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This can be done using only the stuff you say. From there it is fairly easy to compute $BA$ and see that it equals $I$.

Now this is a proof, the only thing unsatisfactory about it is that your very smart 'guess' of the right form of $B$ seems to come out of nowhere. I could write a bit about how you could have found it, but perhaps that is not necessary?

Vincent
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  • Ah, the answer to the question Math_QED posted a link to while I was typing is much nicer! Go check it out. – Vincent Jun 15 '17 at 07:43
  • What if B is not invertible? –  Jun 15 '17 at 07:43
  • Then it is impossible to have $AB = I$. – Vincent Jun 15 '17 at 07:44
  • @Vincent, I don't know/not allowed any of that stuff mentioned that link – Vikram Jun 15 '17 at 07:45
  • Well then maybe my answer is useful after all – Vincent Jun 15 '17 at 07:46
  • very helpful ! thanx – Vikram Jun 15 '17 at 07:47
  • So you say that you have A,B are regular matrices right? Why not use the following argument: $AB = I \Rightarrow BAB = B \Rightarrow BA = I$? –  Jun 15 '17 at 07:47
  • @Math_QED I like it, that is why I said the answer in you link is much nicer than my answer. But I guess for the OP the problem is showing the lemma underlying the last implication namely: let $B$ be given. If there is some $A$ such that $AB = I$ then there is also some $C$ such that $BC = I$. The answer you linked proves this lemma really nicely (in somewhat different language). But it is only after you know this that you can use the clever trick from your last comment to show that actually one can take $C = A$. – Vincent Jun 15 '17 at 07:52
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    Ah that explains a lot! –  Jun 15 '17 at 07:59
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$AB=I \implies rg(Ab)=n \implies rg(A)=n=rg(B) \implies A,B$ invertible for dimension reasons $\implies ABA=IA \iff ABA=AI \iff A^{-1}ABA=BA=A^{-1}AI=I$ where we used that I commutes with A. Alternatively you can do the same thing for $B$.

Either way, it is important here that $A$ and $B$ are linear maps between the same finite dimensional vector space in order for the rank argument to work (otherwise there might still be a kernel although of full rank) and that we have that by using this actually $AB=I$ is sufficient to show that $A$ or $B$ is invertible, which is the really essential part, since then the rest is just a standard proof from group theory!

Felix
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