proof of Taylor’s formula for matrix functions


Let p be a polynomialPlanetmathPlanetmath and suppose A and B are squared matrices of the same size, then p(A+B)=k=0n1k!p(k)(A)Bk where n=deg(p).


Since p is a polynomial, we can apply the Taylor expansionMathworldPlanetmath:


where n=deg(p). Now let x=𝐀+𝐁 and x0=𝐀.

The Taylor expansion can be checked as follows: let p(x)=k=0nakxk for coefficients ak (note that this coefficients can be taken from the space of square matrices defined over a field). We define the formal derivative of this polynomial as p(1)(x)=dpdx=k=1nakkxk-1 and we define p(k)=dp(k-1)dx.

Then p(k)(x)=i=knaii!(i-k)!xi-k and we have 1k!p(k)(x0)=i=knaii!(i-k)!k!(x0)i-k. Now consider

= i=0nai(x0)i+i=1naii(x0)i-1(x-x0)++i=jnaii!(i-j)!j!(x0)i-j(x-x0)j++an(x-x0)n
= a0+a1(x)++ai(j=0ii!(i-j)!j!(x0)i-j(x-x0)j)++an(j=0nn!(n-j)!j!(x0)n-j(x-x0)j)
= k=0nakxi=p(x)

since j=0ii!(i-j)!j!(x0)i-j(x-x0)j=(x)i. ∎

Title proof of Taylor’s formula for matrix functions
Canonical name ProofOfTaylorsFormulaForMatrixFunctions
Date of creation 2013-03-22 17:57:04
Last modified on 2013-03-22 17:57:04
Owner joen235 (18354)
Last modified by joen235 (18354)
Numerical id 4
Author joen235 (18354)
Entry type Proof
Classification msc 47A56