...with Mathematica? Or is this going to be impossible? I don't want to start work if it is hopeless. I've already solved the problem on a smaller scale, which involved an 8x8 matrix. Unfortunately, because of the nature of this problem, the next length scale up jumps to 512x512. Crazy.
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1 Answer
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This should relatively easy to find out.
Make a random Matrix, find out if it's even possible to diagonalize it.
matA = RandomInteger[{-5, 5}, {10, 10}];
DiagonalizableMatrixQ[matA]
*True*
Calculate the Eigenvectors of the matrix to use for further calculations.
$$A_{\text{diag}}={T^{-1}} .A.T$$
matB = Transpose[Eigenvectors[matA]]
matD = Inverse[matB].matA.matB;
DiagonalMatrixQ[matD]
*True*
Assuming your matrix is diagonalisable to begin with, this should give you the solution you're looking for
answered Jul 27, 2019 at 15:16
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$\begingroup$ Good point. I can do a simple test. I'm afraid it might overload my desktop mac. So I'll close all the other programs and I'll be ready to turn the power off if it gets stuck. $\endgroup$– ChrisCommented Jul 27, 2019 at 19:20
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$\begingroup$ Finding the solution of the matrix is diagonalizable or not with a 512x512 matrix only took me a second or two, however i couldnt generate one randomly that was to test with. If you have one already it shouldnt take tooooo much time. $\endgroup$ Commented Jul 27, 2019 at 19:27
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$\begingroup$ I would think that generating the random 512x512 matrix would be the easy part. $\endgroup$– ChrisCommented Jul 27, 2019 at 19:31
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2$\begingroup$ Generating a random one yes...one that is diagonalisable right away no so quick... $\endgroup$ Commented Jul 27, 2019 at 19:33
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4$\begingroup$ @Chris, if your matrix is e.g. Hermitian or normal, then it is guaranteed to be diagonalizable. As you have not mentioned where your matrix came from, we are only able to speculate. $\endgroup$ Commented Jul 28, 2019 at 2:42
Eigensystem? $\endgroup$Eigensystemif you want the exact result (I.e. get back fractions not decimals) that could be memory intensive. $\endgroup$