# Have spd $(A^TA)$ and $(B^TB)$, need $A^TB$.

Given two symmetric positive definite matrices $(A^TA)$ and $(B^TB)$ I need to compute $A^TB$.
$A$ and $B$ are not given directly.
$(A^TA)$ and $(B^TB)$ have the same dimensions. $A$ and $B$ are assumed to have the same dimensions, too.

Is there a way to achieve this?

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If $M$ is an orthogonal matrix ($M^TM=I$) then $(MA)^TMA=A^TA$ but $(MA)^TB=A^TM^TB\neq A^TB$ in general. Hence you can't determine $A^TB$. –  user8268 Apr 24 '12 at 14:49
Could you be bothered to write out whatever "spd" stands for? –  Henning Makholm Apr 24 '12 at 14:54
Thanks, user8268. Would you like to make this an answer instead of a comment? –  Rafael Spring Apr 24 '12 at 15:11

Use Cholesky decomposition to find $L$ such that $L^TL= (A^TA)$, similarly for $M$ such that $M^TM=B^TB$, then compute $L^TM$. Probably this is not what you want.

You may also find the square root of $A^TA$ and $B^TB$, then do the muliplication.

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No. For example, $A,B$ could be unitaries, i.e. $A^TA=I$, $B^TB=I$, and so you have absolutely no information whatsoever on $A^TB$.

A couple examples:

1) $A=B=I$; then $A^TA=B^TB=A^TB=I$.

2) $A=\begin{bmatrix}0&1\\ 1&0\end{bmatrix}$, $B=I$. Then $A^TA=B^TB=I$, but $A^TB=A$.

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