# Thread: Non-linear matrix equation - help!

1. Hi folks,

I have been bouncing my head off an algebraic problem, and it's become quite a headache.

The equation is where , and are all matrices (typically 4*4), and are given and may be complex, is the unknown to be solved for (will usually be required to be real), and is an arbitrary nonzero scalar, I mean it can be any nonzero value that will make the two sides equal - in fact, the equation is a proportion.

I don't have a clue how to begin solving this. Even wrote it out term by term as sixteen huge scalar equations, but this didn't get me any closer.

Any ideas?  2.

3. I'd suggest starting with 2x2 matrices first, then generalizing (or at least expanding). Other than that though, I'm not sure where to go next.  4. Originally Posted by MagiMaster
I'd suggest starting with 2x2 matrices first, then generalizing (or at least expanding). Other than that though, I'm not sure where to go next.
Yes, one colleague over here has just suggested the same, and we are already working on this. With fingers heavily crossed.

Leszek.  5. ////  6. Originally Posted by Leszek Luchowski
Hi folks,

I have been bouncing my head off an algebraic problem, and it's become quite a headache.

The equation is where , and are all matrices (typically 4*4), and are given and may be complex, is the unknown to be solved for (will usually be required to be real), and is an arbitrary nonzero scalar, I mean it can be any nonzero value that will make the two sides equal - in fact, the equation is a proportion.

I don't have a clue how to begin solving this. Even wrote it out term by term as sixteen huge scalar equations, but this didn't get me any closer.

Any ideas?
If you can solve the equation with c=1 then you can generate a solution for any non-zero c by looking at . And if c=0 then X=0 is a solution, so I think you concentrate on the case c=1. Or if det (b) is not ) you could concentrate on the case det(B)=1.

The scalar case is, of course easily solved. and if you take determinants you find in the general case an analagous necessary condition with the obviates caveate if |A|=0.

I don't see a general solution at the moment.  7. Originally Posted by Leszek Luchowski
...
The equation is ...
Any ideas?
You may consider defining the matrix, or Just a thought based on some rough understanding of Lie algebra.

Plus or minus a negative sign,   8. Originally Posted by ought Originally Posted by Leszek Luchowski
...
The equation is ...
Any ideas?
You may consider defining the matrix, or Just a thought based on some rough understanding of Lie algebra.

Plus or minus a negative sign, I was under the impression that stood for the transpose of the matrix X, and T was not itself a matrix. That makes the expression a bit problematic. Also since the exponential of a matrix is of necessity invertible, that constrains what can be an exponential.  9. Originally Posted by DrRocket
I was under the impression that stood for the transpose of the matrix X, and T was not itself a matrix.
Precisely. Thank you for clearing that up. Yes it is the transpose.  10. Originally Posted by DrRocket
I was under the impression that stood for the transpose of the matrix X, and T was not itself a matrix. That makes the expression a bit problematic.
I should have picked a different symbol for the matrix T, to avoid confusion.  11. Leszek, I should clean up some mathematical errors in my above post, but I'm really too new at Lie algebra to do so. It deals with square rotation matrices as a group field over a differential manifold, so I don't know how much of this applies to other square matrices. You could still consider exponents in the mean time. You Taylor expand the the exponent into powers of the exponent. The Euler equation, given the proper choice of exponent, can be regrouped into sines and cosines.  12. Originally Posted by ought
Leszek, I should clean up some mathematical errors in my above post, but I'm really too new at Lie algebra to do so. It deals with square rotation matrices as a group field over a differential manifold, so I don't know how much of this applies to other square matrices. You could still consider exponents in the mean time. You Taylor expand the the exponent into powers of the exponent. The Euler equation, given the proper choice of exponent, can be regrouped into sines and cosines.
I don't know why you think this is Lie algebra. Perhaps it is, but you said so, I didn't because I don't know what Lie algebra is. I am equally confused why you mention exponents. The letter T might look like an exponent, but it merely represents the transpose.

Cheers, Leszek.  13. Originally Posted by ought
Leszek, I should clean up some mathematical errors in my above post, but I'm really too new at Lie algebra to do so. It deals with square rotation matrices as a group field over a differential manifold,
Hi ought, welcome to the forum. I don't think this is quite right. First, I have never heard the expression "group field" - could you explain what you mean?

Second, Lie algebras do not "deal" with any sort of continuous transformation; that's what the group does when it is realized as an action on a manifold or a vector space: the latter realization is called a representation, btw.

Third, rotations are not the only continuous transformations that are realizations of an arbitrary Lie group. You are probably thinking of .
You could still consider exponents in the mean time. You Taylor expand the the exponent into powers of the exponent. The Euler equation, given the proper choice of exponent, can be regrouped into sines and cosines.
This last sentence is most definitely only true for , otherwise all bets are off!. P.S. by edit: I just figured that this is wrong: The groups admit of a similar realization. Sorry about that

Moreover, you have skated over the irritating detail that, in order to exponentiate a matrix, you must first bring it into diagonal form. This, as I am sure you know, is an extremely tedious business, but having made the effort, the rewards are immense.

I hope you can follow this. Challenge: what is one of the principal rewards of bringing an arbitrary transformation matrix to diagonal form?

Leszek: I apologize for seeming to hijack your thread. Truth is, I don't see a general solution. Do we have enough information, I wonder?  14. Originally Posted by Guitarist
Leszek: I apologize for seeming to hijack your thread. Truth is, I don't see a general solution. Do we have enough information, I wonder?
One post is not hijacking. The thread seems to be drifting off-topic because ought has been reading even more complexity into my problem than there was.

Please let me know what information you think would help. I have nothing to hide; not even my credit card number because I don't have one.

Cheers, Leszek.  15. Originally Posted by Guitarist Originally Posted by ought
Leszek, I should clean up some mathematical errors in my above post, but I'm really too new at Lie algebra to do so. It deals with square rotation matrices as a group field over a differential manifold,
Hi ought, welcome to the forum. I don't think this is quite right. First, I have never heard the expression "group field" - could you explain what you mean?
A field consisting of group elements.

Second, Lie algebras do not "deal" with any sort of continuous transformation
Here's one.    I think this is enough spam for now. Later. Much later.  Bookmarks
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