Re: [Algorithms] solving for multiple matrices

[Gino v. d. B. <gin...@gm...>]

If Y can be considered a rotation then Y is orthogonal and thus Y^-1 = 
Y^T,  in which case this equation can be solved through Cholesky 
decomposition:

For C = Y^T * D * Y, let's decompose

C = U^T * U, and
D = V^T * V

then

U^T U = Y^T * V^T * V * Y
            = (V * Y) ^T * (V * Y)

this gives U = V * Y

and thus Y = V^-1 * U

Basically you are taking the square root of a matrix.

Hope this helps,

Gino 



Andras Balogh wrote:
> Both X and Y matrices represent a simple translation and rotation. For the  
> case of the Y matrix, the translation part will likely to be very small,  
> so I could probably pretend it's only rotation.
>
> What I would really like though, is to find a solution, where I could use  
> more than two equations, eg:
> A1 = X * B1 * Y
> A2 = X * B2 * Y
> A3 = X * B3 * Y
> ...
> An = X * Bn * Y
>
> And then compute a least squares solution from this over-constrained  
> system.
> BTW, when I said that I'm looking for an analytical solution, I just meant  
> something that is not based on an iterative approach. As long as I can get  
> to a part where I have to solve a large system of over-constrained linear  
> equations, I'm home. Unfortunately, I don't know how to make this linear..
>
>
> Andras
>
>
>
>
> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:
>
>   
>> Andras Balogh wrote:
>>     
>>> 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?
>>>
>>>
>>>       
>> That's the formula for applying a rotation in the reference frame of
>> another rotation.
>>
>> Do you know anything more about these matrices than that they are
>> matrices? Are they supposed to contain no scale? No translation? If you
>> can formulate them as quaternions, writing out the analytical answer is
>> a lot simpler :-)
>>
>> Sincerely,
>>
>> jw
>>
>>
>>
>>     
>
>
>
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