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If $a \in S$ is some invertible element in a ring $S$, then a computation shows

$$\pmatrix{a & 0 \\ 0 & a^{-1}} = \pmatrix{1 & a \\ 0 & 1} \pmatrix{1 & 0 \\ -a^{-1} & 1} \pmatrix{1 & a \\ 0 & 1} \pmatrix{0 & -1 \\ 1 & 0}.$$

If $R \to S$ is a surjective homomorphism, we see that this invertible matrix over $S$ may be lifted to some invertible matrix over $R$. This observation is important in algebraic K-Theory; for example it is used in the exactness of the relative $K_0$-sequence (see Rosenberg's book, 1.5.4 - 1.5.5).

Questions. What is the idea behind this factorization? Of course it makes no problem to verify this identity, but how can you come up with such a nontrivial factorization? Does it have a geometric interpretation? Who was the first one to find and use this identity?

PS: Isn't it sad that only few textbooks and papers offer explanations of the important insights, rather than only proof verifications?

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I have no answer about the history or geometric interpretation, but this is how I would come up with a similar identity in a more or less algorithmic fashion. The goal is to express


as a product of elementary matrices. To do this it suffices to "reduce it to $I$ elementary row and column operations". I can do that as follows:

  1. Add $a$ times column 2 to column 1, getting $$\left(\begin{array}{cc}a&0\\1&a^{-1}\end{array}\right).$$

  2. Add $1-a^{-1}$ times column 1 to column 2, getting $$\left(\begin{array}{cc}a&a-1\\1&1\end{array}\right).$$

  3. Subtract column 2 from column 1, getting $$\left(\begin{array}{cc}1&a-1\\0&1\end{array}\right).$$

  4. Subtract $a-1$ times column 1 from column 2, getting $I$.

Working backwords, I find the factorisation $$\left(\begin{array}{cc}a&0\\0&a^{-1}\end{array}\right) = \left(\begin{array}{cc}1&a-1\\0&1\end{array}\right)\left(\begin{array}{cc}1&0\\1&1\end{array}\right)\left(\begin{array}{cc}1&a^{-1}-1\\0&1\end{array}\right)\left(\begin{array}{cc}1&0\\-a&1\end{array}\right).$$

Unless I completely misunderstood the motivation, I think this factorisation is just as useful.

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I know this deduction. Can you motivate each step, and their order? – Martin Brandenburg Apr 11 '13 at 16:41
All I'm doing is messing around in the bottom row to create a 1, and then use that 1 to destroy everything else in the bottom row, and then basically I'm done. – Sean Eberhard Apr 11 '13 at 16:45

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