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→Jordan decomposition: link to spectral radius article |
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The above Taylor power series allows the scalar <math>x</math> to be replaced by the matrix. This is not true in general when expanding in terms of <math>A(\eta) = A+\eta B</math> about <math>\eta = 0</math> unless <math>[A,B]=0</math>. A counterexample is <math>f(x) = x^{3}</math>, which has a finite length Taylor series. We compute this in two ways,
* Distributive law: <math display="block">f(A + \eta B) = (A+\eta B)^{3} = A^{3} + \eta(A^{2}B + ABA + BA^{2}) + \eta^{2}(AB^{2} + BAB + B^{2}A) + \eta^{3}B^{3}</math>▼
* Using scalar Taylor expansion for <math>f(a+\eta b)</math> and replacing scalars with matrices at the end
▲:<math>f(A+\eta B) = (A+\eta B)^{3} = A^{3} + \eta(A^{2}B + ABA + BA^{2}) + \eta^{2}(AB^{2} + BAB + B^{2}A) + \eta^{3}B^{3}</math>
▲* Using scalar Taylor expansion for <math>f(a+\eta b)</math> and replacing scalars with matrices at the end :
f(a+\eta b) &= f(a) + f'(a)\frac{\eta b}{1!} + f''(a)\frac{(\eta b)^2}{2!} + f'''(a)\frac{(\eta b)^3}{3!} \\[.5em]
&= a^3 + 3a^2(\eta b) + 3a(\eta b)^2 + (\eta b)^3 \\[.5em]
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