Every square matrix satisfies its own characteristic polynomial if there are no repeated eigenvalues. Δ(A)=0 Δ(λ)=∣λI−A∣=λn+∑i=0n−1ciλi In the case where is diagonalizable (i.e A=PΛP−1 ), Δ(A)=P[Λn+∑i=0n−1ciΛi]P−1. Λn+∑i=0n−1ciΛi is itself a diagonal matrix where the jth entry on the diagonal is λjn+∑i=0n−1ciλj=0 since is a root of the characteristic polynomial. Thus Δ(A)=P⋅0⋅P−1=0 , and −An=∑i=0n−1ciAi.(20) This also gives us a new way to find because by its Taylor series expansion, eAt=∑k=0∞k!1Ak. By Equation 20, all Ak=AnAk−n for can be expressed in terms of the lower powers for . eAt=∑i=0n−1αi(t)Ai for some which are solutions to the equations eλjt=∑i=0n−1αi(t)λji. 
