Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jeffreyH on 03/06/2014 19:22:42
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I have no idea if anyone here can answer this. We have the identity matrix which can be derived from the inverse applied to an original matrix. You can also turn a matrix upside down by selecting the opposite diagonal to the identity matrix. However I have not found a matrix that can produce a transpose of another matrix. Is this possible?
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I'm not sure about the general rule, but certainly for 2x2 matrices you can construct one provided the determinant of the matrix is non-zero. I'll let you figure out how to do it, but it's straightforward to multiply two 2x2 matrices and find one of them which transposes the other...
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I'm not sure about the general rule, but certainly for 2x2 matrices you can construct one provided the determinant of the matrix is non-zero. I'll let you figure out how to do it, but it's straightforward to multiply two 2x2 matrices and find one of them which transposes the other...
I can find the determinant and inverse through transposition, minors and cofactors. That is not the issue. The identity matrix when reversed will invert the position of the rows but I think it is not possible to find a matrix that will transpose another matrix. Doing the transposition by hand or writing an ij -> ji algorithm appears to be only way. I think if a transposing matrix or sequence of matrices could be found it would be very useful for what I am attempting.
Of course once you go past 3 x 3 matrices it is a different story. Not in the sense of transposition or inversion, but of usefulness. Usually x, y and z can be of use with t for time.
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Ok, actually its actually trivial now that I think about it.
We want matrix T and matrix M with transpose of M being MT such that
TM=MT
This is obviously satisfied if T=MTM-1 since
TM=MTM-1M=MT
This requires that M is invertible, which is the determinant requirement I mentioned above.
I think you never see this equation anywhere because computers are far more efficient at just doing the transpose than they are at matrix multiplication.
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Ok, actually its actually trivial now that I think about it.
We want matrix T and matrix M with transpose of M being MT such that
TM=MT
This is obviously satisfied if T=MTM-1 since
TM=MTM-1M=MT
This requires that M is invertible, which is the determinant requirement I mentioned above.
I think you never see this equation anywhere because computers are far more efficient at just doing the transpose than they are at matrix multiplication.
To find T you surely already need to have MT. Now MTM-1 is interesting but not for the reasons you suggest. I haven't worked this through yet for an invertible matrix but will later. If M is not invertible then T could be like a scalar value under certain conditions.
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Actually T can then be considered a scalar transpose. But only if -T11 = -T22 and -T12 = -T21. The initial array M would have the form M11 = -M22 and M12 = -M21
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Ok, actually its actually trivial now that I think about it.
We want matrix T and matrix M with transpose of M being MT such that
TM=MT
This is obviously satisfied if T=MTM-1 since
TM=MTM-1M=MT
This requires that M is invertible, which is the determinant requirement I mentioned above.
I think you never see this equation anywhere because computers are far more efficient at just doing the transpose than they are at matrix multiplication.
Actually I get T as I the identity matrix. So that doesn't work at all. It can't be that T = I otherwise M would have to equal MT
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Ok, actually its actually trivial now that I think about it.
We want matrix T and matrix M with transpose of M being MT such that
TM=MT
This is obviously satisfied if T=MTM-1 since
TM=MTM-1M=MT
This requires that M is invertible, which is the determinant requirement I mentioned above.
I think you never see this equation anywhere because computers are far more efficient at just doing the transpose than they are at matrix multiplication.
Now that I have actually calculated it instead of trying to do it by hand yes that does work and is what I wanted. Thanks JP.
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Interestingly the determinant of T appears to be 1. At least for the example I have used. I wonder if this is a general rule. It is a general rule that the Transposing matrix T has a determinant of 1. I have run tests on a variety of matrices.
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I believe since a transpose doesn't change the eigenvectors, your transposition matrix has to have determinant 1.
This also comes in a straightforward way from the properties of determinants:
det(AB)=det(A)det(B)
and
det(AT)=det(A).
So
det(M)=det(MT)=det(TM)=det(T)det(M),
so det(T)=1.
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I believe since a transpose doesn't change the eigenvectors, your transposition matrix has to have determinant 1.
This also comes in a straightforward way from the properties of determinants:
det(AB)=det(A)det(B)
and
det(AT)=det(A).
So
det(M)=det(MT)=det(TM)=det(T)det(M),
so det(T)=1.
Thanks JP. I have a way to go yet on brushing up on this. It has been quite a while since I have had to use this type of maths. I have to work through various rules and theorems for the work I am doing so may be coming back with a few more questions. I hope you don't mind.
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Also T-1T = I.
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One other thing of note: this all works if you're working with square matrices. This doesn't for for NxM matrices if N and M aren't the same (rectangular matrices). You can't construct an inverse and the whole proof falls apart.
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One other thing of note: this all works if you're working with square matrices. This doesn't for for NxM matrices if N and M aren't the same (rectangular matrices). You can't construct an inverse and the whole proof falls apart.
I had learnt that already :-)