Re: [Algorithms] solving for multiple matrices

[Staffan L. <sta...@ep...>]

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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