3 ms·

Any idea what the derivation in [1] looks like?

• edflsafoiewq 3y ago
  Let α and β be the eigenvalues of the matrix A. A satisfies its characteristic polynomial so we have A^2 - (α + β)A + αβI = 0 If α /= β, define X = (A - βI)/(α - β) Y = (A - αI)/(β - α) which satisfy X^2 = X XY = YX = 0 Y^2 = Y so for k >= 2, X^k = X and Y^k = Y. Hence A^n = (αX + βY)^n = α^n X^n + β^n Y^n = α^n X + β^n Y On the other hand, if α = β, define Z = A - αI which satisfies Z^2 = 0, so Z^k = 0, and A^n = (αI + Z)^n = α^n I + n α^(n-1) Z
  • fweimer 3y ago
    Thanks, that actually looks appropriately elementary.