Diagonalising A = PDP⁻¹ makes Aⁿ easy because Dⁿ raises the diagonal entries to the nth power. Why does that also reveal the long-run behaviour of the system?

Question

Accepted Answer

Because the largest eigenvalue's power eventually dominates

Answer

Because the determinant of Aⁿ approaches zero

Answer

Because all eigenvalues converge to the same value as n grows

Answer

Because P and P⁻¹ cancel, leaving only the initial condition

Diagonalising $A = PDP^{-1}$ makes $A^{n}$ easy because $D^{n}$ raises the diagonal entries to the $n$ th power. Why does that also reveal the long-run behaviour of the system?

Practise Matrix Diagonalisation

More questions on Matrix Diagonalisation

Diagonalising A=PDP−1A = PDP^{-1}A=PDP−1 makes AnA^{n}An easy because DnD^{n}Dn raises the diagonal entries to the nnnth power. Why does that also reveal the long-run behaviour of the system?

Practise Matrix Diagonalisation

More questions on Matrix Diagonalisation

Diagonalising $A = PDP^{-1}$ makes $A^{n}$ easy because $D^{n}$ raises the diagonal entries to the $n$ th power. Why does that also reveal the long-run behaviour of the system?