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I'm trying to understand this proof that every origin-preserving rigid motion can be represented as left-multiplication by a matrix. The proof is given in Artin's "Algebra", pp 127-128. I've posted an image of the relevant text here: enter image description here

The paragraph with the blue line proves that "if m is a map that (1) preserves dot product and (2) fixes the basis vectors, then m is the identity map". That's fine.

In the next paragraph, he assumes that m preserves dot product, builds the matrix A, and asks us to consider the composite map A-1m. It's clear to me that this composite map preserves dot products. But then he declares without justification (underlined in red) that it also fixes the basis vectors. I can't see why.

Since A is orthogonal, A-1 is just AT, right? But even using that, I can't seem to prove that A-1m applied to ei is ei.

What am I missing? Thanks.

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$m$ maps $e_i$ to $e_i'$, by definition of $e_i'$.

$A^{-1}$ maps $e_i'$ to $e_i$, because $A$ maps $e_i$ to $e_i'$, by construction of $A$.

Therefore, $A^{-1}m$ maps $e_i$ to $A^{-1}e_i'$, which is $e_i$.

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Thank you. I see now. –  infinitecardinal Nov 11 '12 at 6:44

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