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- Linear Algebra Foundations #7 - The 1000th Power of a Matrix
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Linear Algebra Foundations #7 - The 1000th Power of a Matrix
Linear Algebra Foundations #7 - The 1000th Power of a Matrix
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What makes this problem extra challenging is that the given matrix M is not diagonalizable. It has a single eigenvalue lambda = 1, and the corresponding eigenspace is unfortunately only one dimensional, spanned by the eigenvector v1 = (-3, 1).
If we can't make M diagonal, we can at least make it upper triangular. Let's grow {v1} to a basis for R^2. For simplicity, let's choose our second basis vector v2 to be orthogonal to v1. I chose v2 = (1, 3). Now just work in this new basis and the calculation works out.
I used the Jordan Decomposition, to find such that . Then found the 1000th power of the Jordan form (which was really simple since the Jordan form was an elementary matrix), then remultiplied the corresponding matrices . Such also factorized really nicely into two elementary matrices which made the final step easier.
Wonder how this method compared to using the Cayley-Hamilton Theorem in terms of speed.
A^1000 can be founded just discovering the pattern
A = -2*-3*(n-1)
B = -9*n
C = n
D = 4+3*n
where n is the nth power of A