0

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?

olgaolgano
  • 55
  • 5
  • That's of course not possible in general for arbitrary $A$ and $B$ (trivial example $A=0, B \neq 0$, but much more can go wrong). For the classical case that $A$ is symmetric and $B$ is diagonal, cf. e.g. https://math.stackexchange.com/q/1382288/96384 and https://math.stackexchange.com/q/1388421/96384 with further links and references therein. – Torsten Schoeneberg Apr 02 '20 at 16:56

1 Answers1

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 $$D=(P_2^{-1})^T\omega^TP_1^T.$$

P. Lawrence
  • 5,674
  • "$A$ and $B$ positive definite" entails in particular they are both symmetric, right? – Torsten Schoeneberg Apr 03 '20 at 00:02
  • @Torsten Schoeneberg : Yes, I assumed that A and B are symmetric. The result I gave for the real positive definite case can easily be generalized to real symmetric matrices having the same rank and signature. It is also true for real symmetric matrices with the same rank over any algebraically closed field of characteristic not 2, in particular over the complex numbers. – P. Lawrence Apr 04 '20 at 05:09
  • Thank you, I guess this is not true over finite fields. Right? – olgaolgano Apr 04 '20 at 14:13
  • 1
    @olgaolgano: It is certainly possible that , for a particular pair of symmetric matrices A,B over a finite field F of chacteristic not equal to 2, all the equations omega_i^2 eta_i =zeta_i could have solutions in F but that would be a very fortuitous occurence. – P. Lawrence Apr 04 '20 at 14:52