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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...

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 thatTM=M^{T} This is obviously satisfied if T=M^{T}M^{-1} sinceTM=M^{T}M^{-1}M=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.

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.

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.