Summary of the problem: Writing the coefficients of the characteristic polynomial of a matrix where we perturb its first column as functions of the coefficients of the characteristic polynomial of the original matrix.
Problem: Let $A \in \mathcal{M}_N(\mathbb{R})$ with $N \in \mathbb{N}^{\star}$ be a real matrix. Consider the coefficients $(\omega_k)_{k=0}^{k = N} \subset \mathbb{R}$ of its characteristic polynomial, thus given by $\chi_{A}(\xi) = \sum_{k = 0}^{k=N} \omega_k \xi^k$. Of course, it is well-known that $\omega_N = 1$ and that the matrix $A$ annihilates is characteristic polynomial $\chi_{A}$, by the Cayley-Hamilton theorem.
Now take $\delta \in \mathbb{R}^N$ and perturb the matrix $A$ on the first column via $\delta$, that is consider $A + \delta \otimes e_1$, where $e_1$ is the first vector of the canonical basis. To be more clear $$ A + \delta \otimes e_1 = \begin{pmatrix} A_{11} + \delta_1 & A_{12} & \dots & A_{1N} \\ A_{21} + \delta_2 & A_{22} & \dots & A_{2N} \\ \vdots & \vdots & & \vdots \\ A_{N1} + \delta_N & A_{N2} & \dots & A_{NN} \\ \end{pmatrix}. $$
Prove that the polynomial $\Phi(\xi) = \sum_{k = 0}^{k = N} \tilde{\omega}_k \xi^k$, where we define $$ \tilde{\omega}_k = \begin{cases} \omega_N, ~ &\text{if} ~ k = N, \\ \omega_k - \left ( \sum_{l = 0}^{N-1-k} \omega_{k + l + 1} A^l (\delta \otimes e_1) \right )_{11}, ~ &\text{if} ~ k = 0, \dots N-1, \\ \end{cases} $$ is an annihilator of $A + \delta \otimes e_1$. Indeed, this proves that $\Phi \equiv \chi_{A + \delta \otimes e_1}$.
Ideas I have experimented with:
- Proceed by recurrence on the dimension $N$ of the problem.
- And/or using the Laplace formula (https://en.wikipedia.org/wiki/Laplace_expansion) for the determinant, knowing that $\chi_A(\xi) = \text{det}(A-\xi I)$.
- Developing the powers of $A + \delta \otimes e_1$ using the Baker-Campbell-Hausdorff formula, see https://mathoverflow.net/questions/78813/binomial-expansion-for-non-commutative-setting
Anyway, I cannot figure out how to solve the problem, even if I know the expression I am looking for.
Origin of the problem: I started from Is it possible to convert a linear system in an ODE of higher order but considering a possibly non linear system of ODEs of the form $y' = Ay + \Delta (y_1)$ where $A \in \mathcal{M}_N(\mathbb{R})$ and $\Delta: \mathbb{R} \to \mathbb{R}^N$ depending only on $y_1$ (the first component of $y \in \mathbb{R}^N$). By recurrence, one case show that for any $k \in \mathbb{N}$ $$ y^{(k)} = A^k y + \sum_{l = 0}^{k - 1}A^{l}(\Delta(y_1))^{(k - 1 - l)}. $$ Using the coefficients of the characteristic polynomial of $A$, we can write $$ \sum_{k = 0}^{N} \omega_k y^{(k)} = \underbrace{\left ( \sum_{k = 0}^{N} \omega_k A^k \right )}_{ = 0}y + \sum_{k = 0}^{N} \omega_k \sum_{l = 0}^{k - 1}A^{l}(\Delta(y_1))^{(k - 1 - l)}, $$ using the Cayley-Hamilton theorem. This yields the equivalent scalar ODE on $y_1$ $$ \sum_{k = 0}^{N} \omega_k y_1^{(k)} - \sum_{k = 0}^{N} \omega_k \sum_{l = 0}^{k - 1}A^{l}(\Delta(y_1))^{(k - 1 - l)} = 0. $$ If, after all this process, we assume that $\Delta$ is a linear function, namely that it exists $\delta$ such that $\Delta(y_1) = \delta y_1 = (\delta \otimes e_1) y$, then we obtain after a change of indices $$ \omega_N y_1^{(N)} + \sum_{k = 0}^{N-1} \left (\omega_k - \sum_{l = 0}^{N-1-k}\omega_{k + l + 1}A^l (\delta \otimes e_1)|_{11} \right )y_1^{(k)} = 0. $$ The aim is to show that the result got if we suppose that the system is fully linear from the very beginning of the reduction from a system to a unique equation on $y_1$ is exactly the same if we suppose linearity at the very end.
Example of application For $N = 3$, take the matrix and the perturbation $$ A = \begin{pmatrix} 1 & 1 & 0\\ 2 & 0 & 1 \\ 0 & 2 & 3 \end{pmatrix}, ~~~~~~ \delta = \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, ~~~~~~ \text{thus} ~~~~~~ A + \delta \otimes e_1 = \begin{pmatrix} 2 & 1 & 0\\ 3 & 0 & 1 \\ 1 & 2 & 3 \end{pmatrix}. $$ Easy computations give $\chi_A(\xi) = \xi^3 - 4\xi^2 - \xi + 8$. Using the definition of characteristic polynomial, we again have $\chi_{A + \delta \otimes e_1} = \xi^3 - 5\xi^2 + \xi + 12$. Using the previous definition
- $k=2$. We have $\tilde{\omega}_2 = \omega_2 - \left ( \sum_{l = 0}^{l = 0} \omega_{3+l} A^l (\delta \otimes e_1) \right )_{11} = -4 - 1 = -5$.
- $k=1$. We have $\tilde{\omega}_1 = \omega_1 - \left ( \sum_{l = 0}^{l = 1} \omega_{2 + l} A^l (\delta \otimes e_1) \right )_{11} = -1 - ((-4) \times 1 + 1 \times 2) = 1$.
- $k=2$. We have that $$ A^2 = \begin{pmatrix} 3 & 1 & 1\\ 2 & 4 & 3\\ 4 & 6 & 11 \end{pmatrix}, $$ thus the usual computations yield $$ \tilde{\omega}_0 = \omega_0 - \left ( \sum_{l = 0}^{2} \omega_{l + 1} A^l (\delta \otimes e_1) \right )_{11} = 8 - ((-1)\times 1 + (-4) \times 2 + 1 \times 5) = 12. $$
Thank you in advance!
Apply the matrix determinant lemma $$ \det(\lambda I-A+uv^T)=\det (\lambda I-A) + v^T \operatorname{adj}(\lambda I-A)u $$ and by Cayley-Hamilton theorem, the adjugate matrix is given by $$ \operatorname{adj}(\lambda I-A)=\sum_{0\leq j+k<N}\omega_{j+k+1}\lambda^j A^k. $$
So $$ \chi_{A+\delta\otimes e_1}(\lambda)=\chi_A(\lambda)-\left[\left(\sum_{0\leq j+k<N}\omega_{j+k+1}\lambda^j A^k\right)\delta\right]_1 =\chi_A(\lambda)-\sum_{j=0}^{N-1}\lambda^j\sum_{k=0}^{N-1-j}\omega_{j+k+1}\left[A^k\delta\right]_1 $$ So extracting the coefficients, $$ \tilde\omega_k=\begin{cases} \omega_N & k=N\\ \omega_k-\sum_{\ell=0}^{N-1-k}\omega_{k+\ell+1}[A^\ell\delta]_1 \end{cases} $$ and of course $[A^\ell\delta]_1=[A^\ell(\delta\otimes e_1)]_{11}$.