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Oct 19, 2014 at 7:43 comment added Federico Poloni (Incidentally, unfortunately one cannot extend these operations to a fully-fledged "transposition algebra" with $T^2=1$, since $A^{1+T}\neq A^{T+1}$, i.e., $A^TA\neq AA^T$.)
Oct 19, 2014 at 7:17 comment added Federico Poloni In numerical linear algebra, where transposes normally go on the right, the notation $A^{-T} := (A^{-1})^T = (A^T)^{-1}$ is quite common, and on rare occasions I have seen $A^{2T}$. So there is an analogously suggestive solution even putting transposes on the right.
Oct 18, 2014 at 23:45 review Low quality posts
Oct 19, 2014 at 0:36
Oct 18, 2014 at 23:25 history answered Bjørn Kjos-Hanssen CC BY-SA 3.0