Does it exist $P,Q \in \mathbb K^n$ such that for all $A \in \mathbb K^n : A^T = PAQ$
I suppose it is not, but how to find a contradiction ? Thank you
Is transpose a change of basis?
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linear-algebra
matrices
duality-theorems
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1Do you mean $M_n(\Bbb K)$? – 2017-02-03
1 Answers
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Suppose $n>1$ of course.
Set $A=I$ to get $Q=P^{-1}$. Since $\phi(A)=PAP^{-1}$ is an automorphism (i.e it satisfies the identity $\phi(AB)=\phi(A)\phi(B)$ for all $A$ and $B$) and $\psi(A)=A^T$ is an antiautomorphism (i.e. it satisfies the identity $\phi(AB)=\phi(B)\phi(A)$ for all $A$ and $B$), the only way they can be equal is if $M_n(\Bbb K)$ is commutative. But whenever $\mathbb{K}$ is a ring with unity this is impossible, since
$$ \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} \ne \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}. $$
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0@Lucas You mean invertible? Setting $A=I$ gives $I=PQ$, so $P$ and $Q$ would have to be inverses (hence invertible). – 2017-02-03
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0@Lucas If $PQ=I$ then $P$ and $Q$ are invertible, then left-multiplying by $P^{-1}$ gives $Q=P^{-1}$ and right-multiplying by $P$ gives $QP=I$. (Not sure why you want $QP=I$ but whatever.) – 2017-02-03