[jeffreyH (OP)]

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?

[JP]

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...

[jeffreyH (OP)]

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.

[JP]

Ok, actually its actually trivial now that I think about it.

We want matrix T and matrix M with transpose of M being M^T such that

TM=M^T

This is obviously satisfied if T=M^TM^-1 since

TM=M^TM^-1M=M^T

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.

[jeffreyH (OP)]

Ok, actually its actually trivial now that I think about it.

We want matrix T and matrix M with transpose of M being M^T such that

TM=M^T

This is obviously satisfied if T=M^TM^-1 since

TM=M^TM^-1M=M^T

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 M^T. Now M^TM^-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.

[jeffreyH (OP)]

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

[jeffreyH (OP)]

Ok, actually its actually trivial now that I think about it.

We want matrix T and matrix M with transpose of M being M^T such that

TM=M^T

This is obviously satisfied if T=M^TM^-1 since

TM=M^TM^-1M=M^T

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 M^T

[jeffreyH (OP)]

Ok, actually its actually trivial now that I think about it.

We want matrix T and matrix M with transpose of M being M^T such that

TM=M^T

This is obviously satisfied if T=M^TM^-1 since

TM=M^TM^-1M=M^T

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.

[jeffreyH (OP)]

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.

[JP]

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(A^T)=det(A).

So

det(M)=det(M^T)=det(TM)=det(T)det(M),

so det(T)=1.

[jeffreyH (OP)]

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(A^T)=det(A).

So

det(M)=det(M^T)=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.

[jeffreyH (OP)]

Also T^-1T = I.

[JP]

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.

[jeffreyH (OP)]

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 :-)