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Cayley-Hamilton

Theorem 16

Every square matrix
AA
satisfies its own characteristic polynomial if there are no repeated eigenvalues.
Δ(A)=0\Delta(A) = 0
Δ(λ)=λIA=λn+i=0n1ciλi\Delta(\lambda) = |\lambda I - A| = \lambda^n + \sum_{i=0}^{n-1} c_i \lambda^i
In the case where
AA
is diagonalizable (i.e
A=PΛP1A = P\Lambda P^{-1}
),
Δ(A)=P[Λn+i=0n1ciΛi]P1.\Delta(A) = P\left[ \Lambda^n + \sum_{i=0}^{n-1}c_i \Lambda^i \right]P^{-1}.
Λn+i=0n1ciΛi\Lambda^n + \sum_{i=0}^{n-1}c_i \Lambda^i
is itself a diagonal matrix where the jth entry on the diagonal is
λjn+i=0n1ciλj=0\lambda_j^n + \sum_{i=0}^{n-1}c_i\lambda_j = 0
since
λj\lambda_j
is a root of the characteristic polynomial. Thus
Δ(A)=P0P1=0\Delta(A) = P \cdot 0 \cdot P^{-1} = 0
, and
An=i=0n1ciAi.(20)-A^n = \sum_{i=0}^{n-1}c_iA^i. \qquad (20)
This also gives us a new way to find
eAte^{At}
because by its Taylor series expansion,
eAt=k=01k!Ak.e^{At} = \sum_{k=0}^{\infty} \frac{1}{k!}A^k.
By Equation 20, all
Ak=AnAknA^k = A^{n}A^{k-n}
for
k>nk>n
can be expressed in terms of the lower powers
AiA^i
for
i[0,n)i\in[0, n)
.

Theorem 17

eAt=i=0n1αi(t)Aie^{At} = \sum_{i=0}^{n-1}\alpha_i(t)A^i
for some
αi\alpha_i
which are solutions to the equations
eλjt=i=0n1αi(t)λji.e^{\lambda_jt} = \sum_{i=0}^{n-1}\alpha_i(t)\lambda_j^i.