# Identity between characteristic polynomial of matrix and its derivative (using the determinant lemma)

Let $$A = \text{diag}(\lambda_1,\dots,\lambda_n)$$ be a diagonal $$n \times n$$ matrix and $$J:=\textbf{1}\textbf{1}^T$$ be the all-ones square matrix. Using the determinant lemma, we can write

$$\det(xI-A-J)=\det(xI-A)\cdot(1-\textbf{1}^T(xI-A)^{-1}\textbf{1})$$

Let $$p(x):=\det(xI-A)$$ be the characteristic polynomial of $$A$$. What can I do to build a relation (equality) between $$p$$ and $$p'$$ (the first derivative of $$p$$). Concretely, I am looking to show

$$\det(xI-A-J)=p(x)-p'(x)$$

which thanks to the lemma is equivalent to showing

$$p'(x)=p(x)\cdot\sum\limits_{i=1}^n\frac{1}{x-\lambda_i}$$

but I don't know how to tackle this. I cannot compare coefficients because a priori the RHS is not a polynomial. Any ideas/directions I can go?

Write polynomial as $$p(x) = a(x-x_1)^{n_1}(x-x_2)^{n_2}\cdots (x-x_k)^{n_k}$$ Now take it natural logarithm:
$$\ln p(x) = \ln a +n_1\ln (x-x_1) + n_2\ln (x-x_2)+\cdots + n _k\ln (x-x_k)$$
• If $p$ is a characteristic polynomial, why $a \neq 1$? Dec 19, 2020 at 12:51