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$\newcommand\la\lambda$

1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rule, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.

2. So, the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j\ge m_j$ for each $j$. So, for some polynomial $R(z)$, $$Q(z)=R(z)\prod_j(z-\la_j)^{n_j}=R_1(z)\prod_j(z-\la_j)^{m_j} =R_1(z)C(z),$$ where $R_1(z):=\prod_j(z-\la_j)^{n_j-m_j}R(z)$ is a polynomial and $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$. So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.

(The previous answer, referred to in your question, does not even mention indices.)

In response to the requests for details in a comment by the OP:

1. Why is $n_j\ge m_j$? How can we know this for certain?

This follows from

Lemma 1: Let $q(z)$ be a polynomial such that $q^{(k)}(\la)=0$ for some $\la$ and all $k=0,\dots,m-1$, where $m$ is an integer $\ge1$. Then $\la$ is a root of $q(z)$ of a multiplicity $n\ge m$ -- that is,  $q(z)=(z-\la)^m S(z)$ for some polynomial $S(z)$ (which may or may not have $\la$ as a root, of some multiplicity).

Proof of Lemma 1: By shifting, without loss of generality $\la=0$. Divide $q(z)$ by $z^m$ with a remainder $r(z)$ of degree $\le m-1$, so that $q(z)=z^m s(z)+r(z)$ for some polynomial $s(z)$. Then for all $k=0,\dots,m-1$ we have $0=q^{(k)}(0)=r^{(k)}(0)$. Because $r(z)$ is of degree $\le m-1$, it follows that $r(z)$ is the zero polynomial, so that $q(z)=z^m s(z)$, which completes the proof of Lemma 1. $\quad\Box$

2. We obtain that $\la_j$ are the roots of $Q(z)$ but why do we need to show that the derivatives of $Q(t)$ equal 0 at each $\la_j$ for this? Is showing that $Q(\la_j)=0$ for each $j$ not enough?

Not in general. If, say, we had the following for some $j$: $\la_j=0$, $m_j\ge2$, and $Q(z)=zQ_1(z)$ for some polynomial $Q_1(z)$ with $Q_1(0)\ne0$, then we would have $Q(0)=0$ but the multiplicity of the root $0$ of $Q(z)$ would be $1\not\ge m_j$.

3. Where does $R(z)$ come from? Since we know the roots and multiplicities of $Q(z)$, shouldn't $R(z)$ not be a constant instead of a polynomial?

Whether $R(z)$ is a constant or not is irrelevant to the proof -- we just need $R(z)$ to be a polynomial. Anyhow, if we only know that the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j$ for each $j$, then $Q(z)$ can be $R(z)\prod_j(z-\la_j)^{n_j}$ for any polynomial $R(z)$ whatsoever.

$\newcommand\la\lambda$

1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rule, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.

2. So, for some polynomial $R(z)$, \begin{equation} Q(z)=R(z)\prod_j(z-\la_j)^{m_j} =R(z)C(z), \tag{1}\label{1} \end{equation} where $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$ (see details on the first equality in \eqref{1} below). So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.

(The previous answer, referred to in your question, does not even mention indices.)

Details on the first equality in \eqref{1}:

Lemma 1: Let $q(z)$ be a polynomial such that $q^{(k)}(\la)=0$ for some $\la$ and all $k=0,\dots,m-1$, where $m$ is an integer $\ge1$. Then $(z-\la)^m$ is a divisor of the polynomial $q(z)$ -- that is,  $q(z)=(z-\la)^m S(z)$ for some polynomial $S(z)$.

Proof of Lemma 1: By shifting, without loss of generality $\la=0$. Divide $q(z)$ by $z^m$ with a remainder $r(z)$ of degree $\le m-1$, so that $q(z)=z^m s(z)+r(z)$ for some polynomial $s(z)$. Then for all $k=0,\dots,m-1$ we have $0=q^{(k)}(0)=r^{(k)}(0)$. Because $r(z)$ is of degree $\le m-1$, it follows that $r(z)$ is the zero polynomial, so that $q(z)=z^m s(z)$, which completes the proof of Lemma 1. $\quad\Box$

Now, by part 1 of the answer, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$. So, for each $j$, by Lemma 1, the polynomial $(z-\la_j)^{m_j}$ is a divisor of the polynomial $Q(z)$. Also, the polynomials $(z-\la_j)^{m_j}$ are coprime for different values of $j$, since the $\la_j$'s are distinct. So, $\prod_j(z-\la_j)^{m_j}$ is a divisor of $Q(z)$; that is, the first equality in \eqref{1} holds.

It remains to provide

$\newcommand\la\lambda$

1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rule, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.

2. So, the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j\ge m_j$ for each $j$. So, for some polynomial $R(z)$, $$Q(z)=R(z)\prod_j(z-\la_j)^{n_j}=R_1(z)\prod_j(z-\la_j)^{m_j} =R_1(z)C(z),$$ where $R_1(z):=\prod_j(z-\la_j)^{n_j-m_j}R(z)$ is a polynomial and $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$. So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.

(The previous answer, referred to in your question, does not even mention indices.)

$\newcommand\la\lambda$

1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rule, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.

2. So, the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j\ge m_j$ for each $j$. So, for some polynomial $R(z)$, $$Q(z)=R(z)\prod_j(z-\la_j)^{n_j}=R_1(z)\prod_j(z-\la_j)^{m_j} =R_1(z)C(z),$$ where $R_1(z):=\prod_j(z-\la_j)^{n_j-m_j}R(z)$ is a polynomial and $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$. So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.

(The previous answer, referred to in your question, does not even mention indices.)

In response to the requests for details in a comment by the OP:

1. Why is $n_j\ge m_j$? How can we know this for certain?

This follows from

Lemma 1: Let $q(z)$ be a polynomial such that $q^{(k)}(\la)=0$ for some $\la$ and all $k=0,\dots,m-1$, where $m$ is an integer $\ge1$. Then $\la$ is a root of $q(z)$ of a multiplicity $n\ge m$ -- that is, $q(z)=(z-\la)^m S(z)$ for some polynomial $S(z)$ (which may or may not have $\la$ as a root, of some multiplicity).

Proof of Lemma 1: By shifting, without loss of generality $\la=0$. Divide $q(z)$ by $z^m$ with a remainder $r(z)$ of degree $\le m-1$, so that $q(z)=z^m s(z)+r(z)$ for some polynomial $s(z)$. Then for all $k=0,\dots,m-1$ we have $0=q^{(k)}(0)=r^{(k)}(0)$. Because $r(z)$ is of degree $\le m-1$, it follows that $r(z)$ is the zero polynomial, so that $q(z)=z^m s(z)$, which completes the proof of Lemma 1. $\quad\Box$

2. We obtain that $\la_j$ are the roots of $Q(z)$ but why do we need to show that the derivatives of $Q(t)$ equal 0 at each $\la_j$ for this? Is showing that $Q(\la_j)=0$ for each $j$ not enough?

Not in general. If, say, we had the following for some $j$: $\la_j=0$, $m_j\ge2$, and $Q(z)=zQ_1(z)$ for some polynomial $Q_1(z)$ with $Q_1(0)\ne0$, then we would have $Q(0)=0$ but the multiplicity of the root $0$ of $Q(z)$ would be $1\not\ge m_j$.

3. Where does $R(z)$ come from? Since we know the roots and multiplicities of $Q(z)$, shouldn't $R(z)$ not be a constant instead of a polynomial?

Whether $R(z)$ is a constant or not is irrelevant to the proof -- we just need $R(z)$ to be a polynomial. Anyhow, if we only know that the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j$ for each $j$, then $Q(z)$ can be $R(z)\prod_j(z-\la_j)^{n_j}$ for any polynomial $R(z)$ whatsoever.

$\newcommand\la\lambda$

1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rulee, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.

2. So, the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j\ge m_j$ for each $j$. So, for some polynomial $R(z)$, $$Q(z)=R(z)\prod_j(z-\la_j)^{n_j}=R_1(z)\prod_j(z-\la_j)^{m_j} =R_1(z)C(z),$$ where $R_1(z):=\prod_j(z-\la_j)^{n_j-m_j}R(z)$ is a polynomial and $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$.

(The previous answer, referred to in your question, does not even mention indices.)

$\newcommand\la\lambda$

1. Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rule, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.

2. So, the $\la_j$'s are roots of $Q(z)$ with respective multiplicities $n_j\ge m_j$ for each $j$. So, for some polynomial $R(z)$, $$Q(z)=R(z)\prod_j(z-\la_j)^{n_j}=R_1(z)\prod_j(z-\la_j)^{m_j} =R_1(z)C(z),$$ where $R_1(z):=\prod_j(z-\la_j)^{n_j-m_j}R(z)$ is a polynomial and $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$. So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.

(The previous answer, referred to in your question, does not even mention indices.)