Let $f:\Bbb C→\Bbb C$ be an analytic function1. Let $\lambda \in\Bbb C$ and let $$ J=\begin{pmatrix} \lambda & 1 & & &\\ &\ddots& \ddots \\ & & \ddots& 1\\ & & &\lambda\end{pmatrix} = \lambda\begin{pmatrix} 1& 0 & & &\\ &\ddots& \ddots \\ & & \ddots& 0\\ & & &1\end{pmatrix} +\begin{pmatrix} 0 & 1 & & &\\ &\ddots& \ddots \\ & & \ddots& 1\\ & & &0\end{pmatrix}=\lambda I+N\in \Bbb C ^{l×l}$$ (Such matrices J appear as Jordan blocks in the Jordan canonical form of a matrix.)Consider the Taylor expansion of f around $\lambda∈\Bbb C$, i.e.,$f(a) =\sum_{k=0}^n\frac{f^{(k)}(\lambda)}{k!}(a−\lambda)^k$.We define a matrix valued function via the formal power series $F(A) =\sum_{k=0}^\infty \frac{f^{(k)}(\lambda)}{k!} (A−\lambda I)k\in\Bbb C^{l×l}$.(By definition,$A^0=I$ for any square matrix A.)One can show that the matrix function F(A) is well-defined, i. e., the above power series converges (absolutely) for any matrix $A\in\Bbb C^{l×l}$, because f is analytic. Establish the following explicit finite formula for the matrix $F(J)\in \Bbb C^{l×l}$ in terms of f and its derivatives in $\lambda$:
$$ F(J) =\begin{pmatrix} f(\lambda) &\frac{f′(\lambda)}{1!} &\frac{f′′(\lambda)}{2!} &\cdots&\frac{f^{(l−1)}(\lambda)}{(l−1)!} &\\ 0 &f(\lambda)& \frac{f′(\lambda)}{1}&\ddots&\vdots\\ 0& 0 & \ddots& \ddots&\frac{f′′(\lambda)}{2!}\\ \vdots &\cdots&\ddots&f(\lambda)& \frac{f′(\lambda)}{1!}\\ 0 &\cdots&\cdots&0&f(\lambda)\end{pmatrix}$$
Answer: To establish the finite formula for the matrix F(J), we first note that J can be expressed as J = A + N, where A is a diagonal matrix with the eigenvalue $\lambda$ and N is a nilpotent matrix $(i.e., N^n = 0)$ as mentioned.
Now, we want to compute the powers of N inductively:
$N^1= N$
$N^2 = N ✕ N = 0$ (since N is nilpotent)
$N^3 = N ✕ N^2 = 0$
...
So, we see that $N^k = 0$ for $ k \ge 2$.
Now, let's compute F(J) using the Taylor expansion:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}]^* (J - \lambda^*I)^k$
Since $N^2 = 0$, the expansion simplifies:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A + N - \lambda I)^k$ Now, we can apply the binomial theorem, which states that for any matrices A and B:
$(A + B)^k = \sum_{j=0}^k (k choose j) A^{(k-j)} B^j$
In our case, $A = A - \lambda I$ and B = N, and k can be any non-negative integer. So, the above expression becomes:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}]$ $\sum_{j=0}^k (k choose j) (A - \lambda I)^{(k-j)} N^j$
Now, since $N^2 = 0$, the terms with $j \ge 2$ in the inner sum vanish, and we're left with:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] [(A - \lambda I)^{(k-0)} N^0] + [\frac{f^{(k)}(\lambda)}{k!}] [(A - \lambda I)^{(k-1)} N^1]$
Since $N^0$ is the identity matrix I, we have:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - λI)^k + \sum_{k=0}^\infty[\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^{(k-1)} N$
Now, we can separate the terms involving A and N:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^k + N \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^{(k-1)}$
The second sum involving N can be simplified using the property of the derivative of the matrix function:
$N * \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^{(k-1)} = N f'(\lambda) \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^{(k-1)}$
Now, we can recognize that the second sum is the derivative of the matrix function with respect to A evaluated at $\lambda$:
$\sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^{(k-1)} = f'(\lambda) ∂F(A)/∂A|_{(\lambda)}$
So, we have:
$F(J) = \sum_{k=0}^\infty [\frac{f^{(k)}(\lambda)}{k!}] (A - \lambda I)^k + N f'(\lambda) ∂F(A)/∂A|_{(λ)}$
Now, the formula for F(J) in terms of f and its derivatives in X is established.
I want to know is my answer correct for this question as i am in the new place , and i don’t know anyone here to ask. Or can anyone provide me hints.
It is true that $N$ is nilpotent, but nilpotent matrix means that for some positive number $k$, we have $N^k =0$ but that does not mean $N^2 = 0$. You can try to work with the following matrix and try to get that $N^2 \geq 0$ but $N^3= 0$.
$$N=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right)$$
We also notice that $N^{n-1} \neq 0$ but $N^n = 0$ when $N$ is $n\times n$ matrix.
Then we try to calculate $F(J)$ since we are substituting $A=J$ so there should not appear any $A$ in the formula, the correct form should be, \begin{align*}F(J) &= \sum_{k=0}^\infty \frac{f^{(k)}(\lambda)}{k!}(J-\lambda I)^k\\&= \sum_{k=0}^\infty \frac{f^{(k)}(\lambda)}{k!}N^k\\ &=\sum_{k=0}^n \frac{f^{(k)}(\lambda)}{k!}N^k \end{align*}
Finally we have some form of finite sum. You should try to think about what shape should $N^k$ looks like and why in the final form matrix if you look it in diagonally every ''layer'' have same coefficient $f^{(k)}$ and how they relate together.