Re: [Algorithms] solving for multiple matrices

[Andras B. <and...@gm...>]

Hah, you're right. I've never realized you could do that. While the  
operations required are indeed trivial, it doesn't seem like I can express  
it with simple matrix operations. So the code to do it will be somewhat  
non-trivial, but that's fine. Thanks!

Andras

On Sat, 19 Sep 2009 08:41:58 -0600, Staffan Langin
<sta...@ep...> wrote:

> Hi there Andras,
>
> C = Y^-1 * D * Y can be linearized by multiplying by Y (from the left).
>
> <=>
>
> Y*C=D*Y
>
> The equation-system, Y*C=D*Y, is trivial to reformulate to the more  
> common
> form Ax=b.
>
>
> Best regards,
>
> Staffan Langin
>
>
>
> -----Original Message-----
> From: Andras Balogh [mailto:and...@gm...]
> Sent: den 17 september 2009 21:38
> To: gda...@li...
> Subject: [Algorithms] solving for multiple matrices
>
> Hi, I have a chain of transformations with multiple unknown (but fixed!)
> transforms. What I do know is the end result transformation and some of
> the transformations in between, and I know these for multiple frames. So
>   from here, I'd like to compute the unknowns. Here it is in more formal
> version:
>
> I'd like to find 2 unknown matrices X and Y. I have 4 known matrices A1,
> A2, B1 and B2, and also know this:
> A1 = X * B1 * Y
> A2 = X * B2 * Y
>
> I can compute X from the first equation:
> X = A1 * Y^-1 * B1^-1
>
> And substitute it into the second:
> A2 = A1 * Y^-1 * B1^-1 * B2 * y
>
> Assigning:
> C = A1^-1 * A2
> D = B1^-1 * B2
>
> Then it becomes:
> C = Y^-1 * D * Y
>
> Now, how do I solve this for Y? This form lookes strangely familiar, but  
> I
> cannot figure out what to do from here (wish I knew how to Google this  
> ;).
> Hopefully there's an analytic solution to this. Any ideas?
>
> Thanks,
>
>
>
> Andras
>
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