Let A be a 7x7 matrix. Find all solutions of the equation A^7 = 0 up to conjugacy.© BrainMass Inc. brainmass.com June 21, 2018, 6:25 am ad1c9bdddf
There is a trick to taking matrices to higher exponents -- diagonalize first.
First we need to take an appropriate non-singular matrix P such that D = P*A*P-1 is diagonal (where P-1 is the inverse of P)
|a 0 0 0 0 . . .|
|0 b 0 0 0 . . .|
|0 0 c 0 0 . . .|
. . . .
There's a way to do this with PT (transpose of P) as well, but that doesn't help because we need a simplification P*P-1 = I
There is some question *if* some A is diagonalizable; the a, b, c have to be independent eigenvalues. However, let's start here by assuming we have seven of them and then we can work back.
A = P-1*D*P
A^7 = P-1*D*P*P-1*D*P*...*P-1*D*P
= P-1*(D^7)*P (noting that P*P-1 = I of course, that's why we did this)
Now P has to be invertible so it cannot have a determinant of zero, so we can't look too far for zeros there.
D^7 is diagonal, with entries a^7, b^7, ...., g^7
Our first thought is to make all these zero, but that is no good because we need seven independent values.
OK, I worked on this direction for a while and got stalled.
Trying back from the other direction: We know A^7 = 0
So A is a root of the equation x^7 = 0
or (x-0)^7 = 0
The minimum polynomial would then be (x - 0)^n, where n is some number from 1 to 7
We seem to be stuck with eigenvalues all equal to zero.
det(tI -A) = t^7 (or t^n where 1<=n<=7)
We can have a matrix with one, two , . . . up to seven t's on the diagonal
Similar matrices will have the same determinant
det (PBP-1) = det P * det B * det P-1 = det P * det B * (1/det P) = det B
Let B be 7 X 7 with any number n of t's on the main diagonal
Let P be any non-singular/invertible 7 X7 matrix
detPBP-1 = det B = t^n
Let PBP-1 = tI - ...
For a 7x7 matrix, the method of finding the solution of A^7 = 0 up to conjugacy is discussed. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.