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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?

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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:

$$ \ln p(x) = \ln a +n_1\ln (x-x_1) + n_2\ln (x-x_2)+\cdots + n _k\ln (x-x_k) $$

Now, calculate the derivative of it...

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  • $\begingroup$ A non-monic characteristic polynomial? $\endgroup$
    – Rodrigo de Azevedo
    Dec 19, 2020 at 12:49
  • $\begingroup$ Don't understand @RodrigodeAzevedo $\endgroup$
    – nonuser
    Dec 19, 2020 at 12:50
  • $\begingroup$ If $p$ is a characteristic polynomial, why $a \neq 1$? $\endgroup$
    – Rodrigo de Azevedo
    Dec 19, 2020 at 12:51
  • $\begingroup$ I never said that....@RodrigodeAzevedo And why do you need that? $\endgroup$
    – nonuser
    Dec 19, 2020 at 12:51
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    $\begingroup$ You did not, but since the OP is interested in characteristic polynomials, this strikes me as excessive generalisation. $\endgroup$
    – Rodrigo de Azevedo
    Dec 19, 2020 at 12:53

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