Finding Matrices that you can multiply wrong, right

Summary

My codegolf.stackexchange post Github Repository with experiments I went on an adventure finding NxN matrices AAA and BBB where AB=10A+B AB = 10A + B AB=10A+B. First of all, for a given candidate A, B is fixed: AB=10A+BA=10AB−1+II=10B−1+A−1B=10(I−A−1)−1B=10(I+(A−I)−1) AB = 10A + B\\ A = 10AB^{-1} + I\\ I = 10B^{-1} + A^{-1}\\ B = 10(I - A^{-1})^{-1}\\ B = 10(I + (A - I)^{-1}) AB=10A+BA=10AB−1+II=10B−1+A−1B=10(I−A−1)−1B=10(I+(A−I)−1) One approach is to take the eigendecomposition of A. From before, A=QΛAQ−1B=10(I−A−1)−1B=10(QIQ−1−QΛA−1Q−1)−1B=Q(10(I−ΛA−1))Q−1 A = Q \Lambda_A Q^{-1}\\ B = 10(I - A^{-1})^{-1}\\ B = 10(Q I Q^{-1} - Q \Lambda_A^{-1} Q^{-1})^{-1}\\ B = Q(\frac{10}{(I - \Lambda_A^{-1})})Q^{-1}\\ A=QΛA​Q−1B=10(I−A−1)−1B=10(QIQ−1−QΛA−1​Q−1)−1B=Q((I−ΛA−1​)10​)Q−1 This shows that A, B share eigenvectors Q, and their eigenvalues are assosciated by ΛAΛB=10ΛA+ΛB \Lambda_A \Lambda_B = 10 \Lambda_A + \Lambda_B\\ ΛA​ΛB​=10ΛA​+ΛB​ Interestingly, this proves that AB = BA, ie our matrices commute! We can calculate the determinant of B as 10N1Πi(1−1λA,i)10^N \frac{1}{\Pi_i (1 - \frac{1}{\lambda_{A,i}})}10NΠi​(1−λA,i​1​)1​ This is very suggestive, but doesn't immediately yield a way to pick a matrix A such that B is small positive integers. Because A and B share eigenvectors, B can be written as a linear combination of (I,A,A2... AN−1)(I, A, A^2 ... ~ A^{N-1})(I,A,A2... AN−1). For example, A=(124113111),B=(746364128)A = \left(\begin{array}{rrr}1 & 2 & 4 \\1 & 1 & 3 \\1 & 1 & 1\end{array}\right), B = \left(\begin{array}{rrr}7 & 4 & 6 \\3 & 6 & 4 \\1 & 2 & 8\end{array}\right)A=⎝⎛​111​211​431​⎠⎞​,B=⎝⎛​731​462​648​⎠⎞​ ΛA=(−6+20006+2000−1),ΛB=(−6+20006+2000−1)\Lambda_A = \left(\begin{array}{rrr}-\sqrt{6} + 2 & 0 & 0 \\0 & \sqrt{6} + 2 & 0 \\0 & 0 & -1\end{array}\right), \Lambda_B = \left(\begin{array}{rrr}-\sqrt{6} + 2 & 0 & 0 \\0 & \sqrt{6} + 2 & 0 \\0 & 0 & -1\end{array}\right)ΛA​=⎝⎛​−6​+200​06​+20​00−1​⎠⎞​,ΛB​=⎝⎛​−6​+200​06​+20​00−1​⎠⎞​ We observe that ΛB=ΛA2−2ΛA+2I\Lambda...