How to find $ P $ such that $ A^\top = PAP^{-1} $?

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Let $ A $ be a matrix such that $A \in \operatorname{Mat}_3(\mathbb{R}) $. How to find $ P \in \operatorname{Mat}_3(\mathbb{R}) $, without doing heavy calculus, such that $ A^\top = PAP^{-1} $, where $ A^\top $ is the transpose of the matrix $ A $. Thanks a lot.

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1 Answer

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Assuming that A is diagonalisable (the algebraic multiplicity of each eigen-value is equal to its geometric multiplicity) then you can find a matrix X (consisting of the eigen-vectors of A) such that: D = $X^{-1}AX$ where D is a diagonal matrix. So then A = $XDX^{-1}$.

So $A^{T}$ = $(XDX^{-1})^{T}$ =$(X^{-1})^{T}D^{T}X^{T}$ =$(X^{T})^{-1}DX^{T}$ $$But D = X^{-1}AX$$ So $A^{T} =Y^{-1}X^{-1}AXY$ where Y = $X^{T}$ so $$A^{T} = (XY)^{-1}AXY$$ So you can take $(XY)^{-1}$ = P

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