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Eigenvalues and eigenvectors of a 3x3 matrix (principal stresses)[programming] |
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| Apr15-10, 02:08 PM | #1 |
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Eigenvalues and eigenvectors of a 3x3 matrix (principal stresses)[programming]
I need to compute the 3 eigenvalues and 3 eigenvectors of a symmetric 3x3 matrix, namely a stress tensor, computationaly (in C++). More specific details here. Basically 2 questions:
1. I am running into trouble in finding the 3 real roots. Here we have the general formula for the roots. According to Wikipedia, again, "The characteristic equation has three real roots λ, i.e. not imaginary due to the symmetry of the stress tensor.", then I'm sure these huge formulas can be simplified, and it is indeed a hard work to try simplifying it. Have you ever seen a simplified formula for cubic equations where the discriminant is always greater than zero (then, three real roots). 2. After finding the eigenvalues how to find the eigenvectors? I read about general purpose methods like the Power Method but these are for nxn matrices. For a 3x3 matrix there must be a much simpler technique. Thanks in advance. |
| Apr15-10, 02:52 PM | #2 |
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You can find the eigenvector belonging to eigenvalue [itex]\lambda[/itex] by solving the system of equations [itex]Tv=\lambda v[/itex]. You will get three equations which you can solve for v_1,v_2 and v_3.
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| Apr15-10, 03:13 PM | #3 |
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| Apr15-10, 08:07 PM | #4 |
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Eigenvalues and eigenvectors of a 3x3 matrix (principal stresses)[programming]
Mathematica would just give you the answer, Eigenvalues[matrix], done. To calculate yourself, sure get your cubic polynomial and figure out a way to solve it, numerically perhaps???
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| Apr15-10, 08:36 PM | #5 |
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There's probably an even easier way for 3x3 matrices, but Jacobi (Givens) rotation is fairly easy to code and should be quick. See http://scholar.lib.vt.edu/theses/ava.../chapter5a.PDF and http://en.wikipedia.org/wiki/Jacobi_...alue_algorithm.
edit: Actually this page is more describing than the other Wikipedia link imho: http://en.wikipedia.org/wiki/Jacobi_rotation |
| Apr15-10, 09:10 PM | #6 |
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-if these two cross products are zero, the rows of (A-λI) are linearly dependent, then the eigenspace is a plane orthogonal to any of the rows of (A-λI) -if one of the cross products is different zero, then the cross product is the eigenvector itself -it should not be possible for both to be different zero because then the eigenspace would be zero dimensional What you think eh??? Thanks |
| Apr15-10, 09:18 PM | #7 |
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| Apr26-10, 12:51 PM | #8 |
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Now I see that its better to use a numerical method, specially because I don't need a very accurate approximation for my specific application. Then I will give a try to that Jacobi method. Do you guys think this is one of the best choices for symmetrical 3x3 matrices?
Thanks. |
| Apr26-10, 01:05 PM | #9 |
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That was my feeling before, that numerically you can solve a cubic polynomial to whatever accuracy you desire, up to some fundamental limit of time cost perhaps, but that you should be able to get really close without too much trouble. However, it is possible to analytically solve it!!! e.g. http://en.wikipedia.org/wiki/Cubic_function
That should be really fast! And exact too! |
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