Solving matrix functions using Laplace transform of Caley-Hamilton theorem

Laplace meets Cayley–Hamilton

Let f be a polynomial of degree k, f(x) = b_k x^k + b_k-1 x^k-1 + ··· + b₁ x + b₀ and A = (a_ij )_n x n . Define
f(A) = b_k A^k + b_k-1 A^k-1 + ··· + b₁ A + b₀ I. If A is diagonalizable, A = TDT ^{- 1} where D = diag (₁ , ..., _n ), then

(1)		f(A)	= bk TDkT - 1 + bk - 1 TDk - 1T - 1 + ··· + b1 TDT - 1 + b0 TT - 1
			= T [ bk Dk + bk - 1 Dk - 1 + ··· + b1 D + b0 I ] T - 1
			= Tf(D)T - 1.

If p is the characteristic polynomial of A, then p(D) = 0, because
D^k = diag (₁ ^k, ..., _n ^k)    , k = 0, 1, ..., n

=>      p(D) =		p(1 )		0		= 0n x n
			·.
		0		p(n )

It now follows from (1) that p(A) = 0 (when A is diagonalizable). More generally,
the Cayley-Hamiltonin theorem: If p is the characteristic polynomial of a square matrix A ,then p (A) = 0, that is, A satisfies its own characteristic equation.
Example 1: Cayley-Hamilton theorem
Let p be the characteristic polynomial of A.
(2)
p() = ( - 1)ⁿⁿ + b_{n - 1} ^{n - 1} + ··· + b₁ + b₀ = 0 (3)
(-1)ⁿAⁿ + b_{n - 1} A^{n - 1} + ··· + b₁ A + b₀ I = 0 =>      Aⁿ = ( - 1)^{n + 1} [ b_{n - 1} A^{n - 1} + ··· + b₁ A + b₀ I ]
=>      A^{n + 1} = ( - 1)^{n + 1} [ b_{n - 1} Aⁿ + ··· + b₁ A² + b₀ A ] where we can substitute Aⁿ from the second last equation. It follows that all the powers A^k, k n, can be written in terms of A⁰, A¹, ..., A^n-1.
Example 2: Powers of a matrix

If A ^{- 1} exists, that is, if = 0 is not an eigenvalue of A => det A = p(0) = b₀ 0, it follows from (3) that
I = - (b₁ / b₀ ) A - ··· - (b_n-1 / b₀ ) A^n-1 - [(-1)ⁿ / b₀ ] Aⁿ, and thus
(4)
A ^-1 = - ( b₁ / b₀ ) I - ( b₂ / b₀ ) A - ··· - ( b_{n - 1} / b₀ ) A^n-2 - [( - 1)ⁿ / b₀ ] A^n-1. Next consider functions f (A) of matrices, where f : C -> C can be expressed as a power series
(5)
f(z) = c_k z^k. Define
(6)
f(A) = c_k A^k         (A⁰ = I) if the series converges. The following result can be shown:
Proposition 1. If the radius of convergence of the power series (5) is r ( f can be expressed as (5) when z, | z | < r ), then the matrix series (6) converges if the spectral radius r (A) of A satisfies r (A) < r. (r(A) = max{| ₁ | , ..., | _n |}).
Since all the powers A^k, k n, can be written in terms of A⁰, A¹, ..., A^n-1, it follows from (6) that
(7)
f(A) = d₀ I + d₁ A + d₂ A² + ··· + d_{n - 1} A^{n - 1}. The coefficients d₀ , ..., d_{n - 1} depend from A. They satisfy the equation
f(_i ) = d₀ + d_{1 i} + d₂ ²_i + ··· + d_{n - 1 i} ^{n - 1},    i = 1, ..., n where _i 's are the eigenvalues of A.
f(_i ) = c_{k i} ^k  , p (_i ) = 0   =>    f(_i ) = d₀ + d_{1 i} + ··· + d_{n - 1 i} ^{n - 1} Example 3: sin A
Note. If q() = 0, where q is a polynomial, it is not necessarily true that q(A) = 0. There exists a polynomial of minimal degree, called the minimal polynomial of A, such that m(A) = 0.
If the multiplicity of an eigenvalue _i is bigger than one, we may also use the equations
f '(_i ) = d₁ + 2d_{2 i} + ··· + (n - 1)d_{n - 1 i} ^{n - 2} If A is diagonalizable, then

f(A)	= ck Ak = ck TDkT - 1 = T [ ck diag (i k, ..., n k) ] T - 1
	= T [ diag (ck i k, ..., ck n k) ] T -1
	= T diag (ck 1 k, ..., ck n k)T - 1
	= T diag (f (1 ), ..., f (n ))T - 1

and thus
(8)
f(A) = T diag (f (₁ ), ..., f (_n ))T ^{- 1}. http://s-mat-pcs.oulu.fi/~mpa/matreng/ematr6.htm

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Laplacing

Another way which can solve for matrix functions, is too take the Laplace transform and solve in an algebraic mapping state, it also works for repeated eigenvector problems, of course being linear and all.
Below are my notes to the solution, I considered the idea after first being introduced to Caley-Hamilton while attending Swinburne University I spent about a week marinating in the idea and considerable time away from this plane of existence finding a solution.

Interestingly I also stumbled across an new way of solving partial fractions using only a matrix operation, more on this later!

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