### -synx-'s blog

By -synx-, history, 13 months ago, ,

Can we find matrix modular inverse as
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 » 13 months ago, # |   0 clearly no?A= 2*I where I is the identityP = 5A^3 = (2*I)^3 = 8*I, while A^-1 = (0.5*I).
•  » » 13 months ago, # ^ |   0 . For scalar matrices it's true: it's just a usual Fermat's Little Theorem.
•  » » » 13 months ago, # ^ |   0 right, sorry my bad for giving such a poor example as i wanted to ease calculation.
•  » » » 13 months ago, # ^ | ← Rev. 2 →   +11 For diagonal matrices it's also true. For diagonalisable invertible matrices, it looks OK, surprisingly. For matrices not diagonalisable, it is in general false. See e.g. [1, 1] [0, 1] To the power a it is [1, a] [0, 1] therefore the inverse is given for the power a=p-1 and not p-2 (yeah it's close, but...)
 » 13 months ago, # | ← Rev. 2 →   +9 For arbitrary matrices it's false. For example, . Mp - 1 = 2(p - 1) / 2E. If 2 isn't a square modulo p, Mp - 1 ≠ E.Also you can use Lagrange's theorem for group. The order of this group is (pn - 1)(pn - p)... (pn - pn - 1). For n = 2 we get A(p2 - 1)(p2 - p) = E. Doesn't look like the best estimation, though.
•  » » 13 months ago, # ^ |   0 Thank you :)
 » 13 months ago, # |   0 The fact that the right hand side always exists and the left hand side doesn't have to exist hints that it shouldn't be so.Look up the Jordan form. Not only does it tell you when it's possible to use FLT, it basically gives you a formula for matrix powers (in , not ). Of course, computing the Jordan form is itself difficult and you're better off just using fast exponentiation.