I am reading a paper for my thesis, and at some point I think the authors changed $(AA^t)^{-1}A\mathbf v$ into $\mathbf v^t A^{-1}$, where $A$ is an invertible square matrix and $\mathbf v$ a column vector.
I tried to show that it was true and I used the fact that $(A^t)^{-1} = (A^{-1})^t$ but haven't been able to prove it.
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https://math.stackexchange.com/q/340233/305862 – Jean Marie Jun 11 '24 at 18:35
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2Does this answer your question? Transpose of inverse vs inverse of transpose – J.G.131 Jun 11 '24 at 18:40
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$(AA^T)^{-1}Av=(A^{T})^{-1} A^{-1}Av=(A^T)^{-1}v$
$(A^T)^{-1}v=(v^TA^{-1})^{T}$
Bowei Tang
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Yes, this is correct, I just noticed that this appears in the paper only under the norm sign, that's why the transpose outside the parenthesis was not included. Thank you! – oxedex Jun 11 '24 at 19:58
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You have $$ (AA^t)^{-1}Av=(A^{-t}A^{-1})Av=A^{-t}(A^{-1}A)v=A^{-t}v. $$ The transpose of this is $$ (A^{-t}v)^t=v^tA^{-1}. $$
Martin Argerami
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