Finding a $3\times 3$ matrix over $\mathbb{F}$ such that $DAD^T=B$

Question

For 2 given matrices in $\mathbb{F}^{3\times 3}$, say $A$ and $B$, what is the easiest way to find a $3\times 3$ matrix over $\mathbb{F}$ such that $DAD^T=B$?. Note that, there is no supposition on $D$ to have its transpose is equal to its inverse, and there is no particular supposition on the given matrices A and B. Ideas?

Answer 1

All the matrices will be $n \times n$matrices over the field $\mathbb F.$Let $P_1$ and $P_2$ be invertible matrices such that $$P_1^TAP_1=\eta =\text { diag }(\eta_1,...,\eta_n) \text { and }$$ $$P_2^TBP_2=\zeta =\text { diag }(\zeta_1,...,\zeta_n) $$ Suppose there are $\omega_1,...,\omega_n \in \mathbb F$such that $\omega_i^2\eta_i=\zeta_i \text { for }i=1,...,n.$[This will be the case if $\mathbb {F=R}$ and both $A$ and $B$are ositive-definite.]Let $\omega= \text {diag}(\omega_1,...,\omega_n ).$ Then $$\omega^T\eta \omega=\zeta$$ $$\omega^TP_1^TAP_1 \omega=P_2^TBP_2$$ $$(P_2^{-1})^T\omega^TP_1^TAP_1 \omega P_2^{-1}=B$$ $$(P_2^{-1})^T\omega^TP_1^TA((P_2^{-1})^T\omega^TP_1^T)^T=B$$ Then $$D=(P_2^{-1})^T\omega^TP_1^T.$$