# Taylor expansion of a matrix to scalar function

Consider a matrix $A$ and its characteristic equation $$\Phi(A) = \sum_{k=0}^n c_k(A) \lambda^{n-k}.$$ $c_k$ can be worked out in many ways but one is a recursive method derived from the Fadddeev-LeVerrier algorithm: $$c_k(A) = -\frac{1}{k}\sum_{i=0}^{k-1} \operatorname{Tr}[A^{k-i}]c_i(A).$$

I would like to Taylor expand $c_k$ around a small perturbation of $A$.

I have read here (https://mathoverflow.net/questions/139643/taylor-expansion-of-a-function-of-a-matrix) that the formula for this to first order is $$f(A+\epsilon B) = f(A) + \epsilon f'_A(B) + \mathcal{O}(\epsilon^2)$$ but I do not know what the notation `$f'_A(B)$' means.

I have tried to apply this to $c_k$ and, guessing that we can use the product rule, get to $$c_k(A+\epsilon B) \simeq c_k(A) -\frac{\epsilon}{k} \sum_{i=0}^{k-1} \operatorname{Tr}[A^{k-i}]'_A(B) c_i(A) + \operatorname{Tr}[A^{k-i}]c_i(A)'_A(B).$$ I have looked online for notes on how to do this but can't find anything.

Questions:

1. Do you have a reference for the equation Taylor expansion equation that explains it?

2. Can you help finish the Taylor expansion of $c_k(A)$?

Thank you.

• I now see that I can expand $(A+\epsilon B)^{k-i}$ with the binomial theorem which to first order will give $A^{k-i}$ plus $(k-i)$ terms of the form $\epsilon A^xBA^y$ where $x+y = k-i-1$. Under cyclicity of the trace this becomes $tr[A^{k-i}] + \epsilon(k-i)tr[A^{k-i-1}H]$. The full result will come about by doing this expansion recursively, without me needing to know more maths. I am unfamiliar with matrix calculus and the notation $Df_A$ where $A$ is a matrix, as given in the answer below. However that is a different question so I'll mark this as answered. Nov 21 '16 at 14:52

Let $H\in M_n$ be a small matrix. Then $c_k(A+H)\approx c_k(A)+Dc_{k}(H)$ where the last term is given by the recurrence formula:
$Dc_0(H)=0;k\geq 1\implies Dc_{k}(H)=-1/k\sum_{i=0}^{k-1}((k-i)tr(A^{k-i-1}H)c_i(A)+tr(A^{k-i})Dc_{i}(H))$.
• Am I right that you claim that $D(tr[A^{k-i}])(H) = (k-i)tr[A^{k-i-1}H]$? What does $D$ mean? Do you have a reference for this so that I can prove this myself? Furthermore do you have a reference for the Taylor expansion that you use? When I say reference the proper name will do and then I can look it up. I can see that my result will be recursive so maybe it isn't a great example. So it is just question 1 that still stands after your answer. Otherwise thank you very much. Aug 5 '16 at 10:01
• I use the Taylor formula until the degree $1$. If $f:A\in\mathbb{R^n}\rightarrow f(A)\in \mathbb{R}^p$, then its derivative: $Df_A:H\in \mathbb{R^n}\rightarrow Df_A(H)\in \mathbb{R}^p$ is a linear application. Read a book about differential calculus.