Is Sum of Principal Minors Equals to Pseudo Determinant? I'd like to prove following statement and check whether it's true or not.

Let $M$ be a diagonalizable $n × n$ matrix. If the rank of $M$ equals $r (> 0)$, then the pseudo determinant pdet$M$ equals the sum of all principal minors of order $r$.

Pseudo determinant refers to the product of all non-zero eigenvalues of a square matrix. Eigenvalues are scaling factors as far as I know. And principal minors of order r, is also small-sized scaling factors(determinant) of given $M$. 
But does pseudo determinant equal to sum of all principal minor? It looks to me multiplication of those equals to pseudo determinant. 
Which one is correct?
 A: In order not to leave this question unanswered, let me prove the claim along
the lines I've suggested in the comments.
Let us agree on a few notations:


*

*Let $n$ and $m$ be two nonnegative integers. Let $A=\left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}$ be an $n\times m$-matrix (over some ring).
Let $U=\left\{  u_{1}<u_{2}<\cdots<u_{p}\right\}  $ be a subset of $\left\{
1,2,\ldots,n\right\}  $, and let $V=\left\{  v_{1}<v_{2}<\cdots<v_{q}\right\}
$ be a subset of $\left\{  1,2,\ldots,m\right\}  $. Then, $A_{U,V}$ shall
denote the submatrix $\left(  a_{u_{i},v_{j}}\right)  _{1\leq i\leq p,\ 1\leq
j\leq q}$ of $A$. (This is the matrix obtained from $A$ by crossing out all
rows except for the rows numbered $u_{1},u_{2},\ldots,u_{p}$ and crossing out
all columns except for the columns numbered $v_{1},v_{2},\ldots,v_{q}$.) For
example,
\begin{equation}
\begin{pmatrix}
a_{1} & a_{2} & a_{3} & a_{4}\\
b_{1} & b_{2} & b_{3} & b_{4}\\
c_{1} & c_{2} & c_{3} & c_{4}\\
d_{1} & d_{2} & d_{3} & d_{4}
\end{pmatrix}
_{\left\{  1,3,4\right\}  ,\left\{  2,4\right\}  }
=
\begin{pmatrix}
a_{2} & a_{4}\\
c_{2} & c_{4}\\
d_{2} & d_{4}
\end{pmatrix} .
\end{equation}

*If $n$ is a nonnegative integer, then $I_n$ will denote the $n\times n$
identity matrix (over whatever ring we are working in).
Fix a nonnegative integer $n$ and a field $\mathbb{F}$.
We shall use the following known fact:

Theorem 1. Let $\mathbb{K}$ be a commutative ring. Let $A$ be an $n\times
n$-matrix over $\mathbb{K}$. Let $x\in\mathbb{K}$. Then,
  \begin{align}
\det\left(  A+xI_n \right)   &  =\sum_{P\subseteq\left\{  1,2,\ldots
,n\right\}  }\det\left(  A_{P,P}\right)  x^{n-\left\vert P\right\vert
}
\label{darij.eq.t1.1}
\tag{1}
\\
&  =\sum_{k=0}^{n}\left(  \sum_{\substack{P\subseteq\left\{  1,2,\ldots
,n\right\}  ;\\\left\vert P\right\vert =n-k}}\det\left(  A_{P,P}\right)
\right)  x^{k}.
\label{darij.eq.t1.2}
\tag{2}
\end{align}

Theorem 1 appears, e.g., as Corollary 6.164 in my Notes on the combinatorial
fundamentals of algebra, in the version of 10th January
2019 (where I
use the more cumbersome notation $\operatorname*{sub}\nolimits_{w\left(
P\right)  }^{w\left(  P\right)  }A$ instead of $A_{P,P}$). $\blacksquare$

Corollary 2. Let $A$ be an $n\times n$-matrix over a field $\mathbb{F}$.
  Let $r\in\left\{  0,1,\ldots,n\right\}  $. Consider the $n\times n$-matrix
  $tI_n +A$ over the polynomial ring $\mathbb{F}\left[  t\right]  $. Its
  determinant $\det\left(  tI_n +A\right)  $ is a polynomial in $\mathbb{F}
\left[  t\right]  $. Then,
  \begin{align}
&  \left(  \text{the sum of all principal }r\times r\text{-minors of }A\right)
\nonumber\\
&  =\left(  \text{the coefficient of }t^{n-r}\text{ in the polynomial }
\det\left(  tI_n +A\right)  \right)  .
\end{align}

Proof of Corollary 2. We have $r\in\left\{  0,1,\ldots,n\right\}  $, thus
$n-r\in\left\{  0,1,\ldots,n\right\}  $. Also, from $tI_n +A=A+tI_n $, we
obtain
\begin{equation}
\det\left(  tI_n +A\right)  =\det\left(  A+tI_n \right)  =\sum_{k=0}
^{n}\left(  \sum_{\substack{P\subseteq\left\{  1,2,\ldots,n\right\}
;\\\left\vert P\right\vert =n-k}}\det\left(  A_{P,P}\right)  \right)  t^{k}
\end{equation}
(by \eqref{darij.eq.t1.2}, applied to $\mathbb{K}=\mathbb{F}\left[  t\right]
$ and $x=t$). Hence, for each $k\in\left\{  0,1,\ldots,n\right\}  $, we have
\begin{align*}
&  \left(  \text{the coefficient of }t^{k}\text{ in the polynomial }
\det\left(  tI_n +A\right)  \right) \\
&  =\sum_{\substack{P\subseteq\left\{  1,2,\ldots,n\right\}  ;\\\left\vert
P\right\vert =n-k}}\det\left(  A_{P,P}\right)  .
\end{align*}
We can apply this to $k=n-r$ (since $n-r\in\left\{  0,1,\ldots,n\right\}  $)
and thus obtain
\begin{align*}
&  \left(  \text{the coefficient of }t^{n-r}\text{ in the polynomial }
\det\left(  tI_n +A\right)  \right) \\
&  =\sum_{\substack{P\subseteq\left\{  1,2,\ldots,n\right\}  ;\\\left\vert
P\right\vert =n-\left(  n-r\right)  }}\det\left(  A_{P,P}\right)
=\sum_{\substack{P\subseteq\left\{  1,2,\ldots,n\right\}  ;\\\left\vert
P\right\vert =r}}\det\left(  A_{P,P}\right)  \qquad\left(  \text{since
}n-\left(  n-r\right)  =r\right) \\
&  =\left(  \text{the sum of all principal }r\times r\text{-minors of
}A\right)
\end{align*}
(by the definition of principal minors). This proves Corollary 2.
$\blacksquare$

Lemma 3. Let $A$ be an $n\times n$-matrix over a field $\mathbb{F}$. Let
  $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ be the eigenvalues of $A$. We
  assume that all $n$ of them lie in $\mathbb{F}$. Then, in the polynomial ring
  $\mathbb{F}\left[  t\right]  $, we have
  \begin{equation}
\det\left(  tI_n +A\right)  =\left(  t+\lambda_{1}\right)  \left(
t+\lambda_{2}\right)  \cdots\left(  t+\lambda_{n}\right)  .
\end{equation}

Proof of Lemma 3. The eigenvalues of $A$ are defined as the roots of the
characteristic polynomial $\det\left(  tI_n -A\right)  $ of $A$. (You may be
used to defining the characteristic polynomial of $A$ as $\det\left(
A-tI_n \right)  $ instead, but this makes no difference: The polynomials
$\det\left(  tI_n -A\right)  $ and $\det\left(  A-tI_n \right)  $ differ
only by a factor of $\left(  -1\right)  ^{n}$ (in fact, we have $\det\left(
A-tI_n \right)  =\left(  -1\right)  ^{n}\det\left(  tI_n -A\right)  $), and
thus have the same roots.)
Also, the characteristic polynomial $\det\left(  tI_n -A\right)  $ of $A$ is
a monic polynomial of degree $n$. And we know that its roots are the
eigenvalues of $A$, which are exactly $\lambda_{1},\lambda_{2},\ldots
,\lambda_{n}$ (with multiplicities). Thus, $\det\left(  tI_n -A\right)  $ is
a monic polynomial of degree $n$ and has roots $\lambda_{1},\lambda_{2}
,\ldots,\lambda_{n}$. Thus,
\begin{equation}
\det\left(  tI_n -A\right)  =\left(  t-\lambda_{1}\right)  \left(
t-\lambda_{2}\right)  \cdots\left(  t-\lambda_{n}\right)
\end{equation}
(because the only monic polynomial of degree $n$ that has roots $\lambda
_{1},\lambda_{2},\ldots,\lambda_{n}$ is $\left(  t-\lambda_{1}\right)  \left(
t-\lambda_{2}\right)  \cdots\left(  t-\lambda_{n}\right)  $). Substituting
$-t$ for $t$ in this equality, we obtain
\begin{align*}
\det\left(  \left(  -t\right)  I_n -A\right)   &  =\left(  -t-\lambda
_{1}\right)  \left(  -t-\lambda_{2}\right)  \cdots\left(  -t-\lambda
_{n}\right)  \\
&  =\prod_{i=1}^{n}\underbrace{\left(  -t-\lambda_{i}\right)  }_{=-\left(
t+\lambda_{i}\right)  }=\prod_{i=1}^{n}\left(  -\left(  t+\lambda_{i}\right)
\right)  \\
&  =\left(  -1\right)  ^{n}\underbrace{\prod_{i=1}^{n}\left(  t+\lambda
_{i}\right)  }_{=\left(  t+\lambda_{1}\right)  \left(  t+\lambda_{2}\right)
\cdots\left(  t+\lambda_{n}\right)  } \\
& = \left(  -1\right)  ^{n}\left(
t+\lambda_{1}\right)  \left(  t+\lambda_{2}\right)  \cdots\left(
t+\lambda_{n}\right)  .
\end{align*}
Comparing this with
\begin{equation}
\det\left(  \underbrace{\left(  -t\right)  I_n -A}_{=-\left(  tI_n 
+A\right)  }\right)  =\det\left(  -\left(  tI_n +A\right)  \right)  =\left(
-1\right)  ^{n}\det\left(  tI_n +A\right)  ,
\end{equation}
we obtain
\begin{equation}
\left(  -1\right)  ^{n}\det\left(  tI_n +A\right)  =\left(  -1\right)
^{n}\left(  t+\lambda_{1}\right)  \left(  t+\lambda_{2}\right)  \cdots\left(
t+\lambda_{n}\right)  .
\end{equation}
We can divide both sides of this equality by $\left(  -1\right)  ^{n}$, and
thus obtain $\det\left(  tI_n +A\right)  =\left(  t+\lambda_{1}\right)
\left(  t+\lambda_{2}\right)  \cdots\left(  t+\lambda_{n}\right)  $. This
proves Lemma 3. $\blacksquare$
Let us also notice a completely trivial fact:

Lemma 4. Let $\mathbb{F}$ be a field. Let $m$ and $k$ be nonnegative
  integers. Let $p\in\mathbb{F}\left[  t\right]  $ be a polynomial. Then,
  \begin{align*}
& \left(  \text{the coefficient of }t^{m+k}\text{ in the polynomial }p\cdot
t^{k}\right)  \\
& =\left(  \text{the coefficient of }t^{m}\text{ in the polynomial }p\right)
.
\end{align*}

Proof of Lemma 4. The coefficients of the polynomial $p\cdot t^{k}$ are
precisely the coefficients of $p$, shifted to the right by $k$ slots. This
yields Lemma 4. $\blacksquare$
Now we can prove your claim:

Theorem 5. Let $A$ be a diagonalizable $n\times n$-matrix over a field
  $\mathbb{F}$. Let $r=\operatorname*{rank}A$. Then,
  \begin{align*}
&  \left(  \text{the product of all nonzero eigenvalues of }A\right)  \\
&  =\left(  \text{the sum of all principal }r\times r\text{-minors of
}A\right)  .
\end{align*}
  (Here, the product of all nonzero eigenvalues takes the multiplicities of the
  eigenvalues into account.)

Proof of Theorem 5. First of all, all $n$ eigenvalues of $A$ belong to
$\mathbb{F}$ (since $A$ is diagonalizable). Moreover, $r=\operatorname*{rank}
A\in\left\{  0,1,\ldots,n\right\}  $ (since $A$ is an $n\times n$-matrix).
The matrix $A$ is diagonalizable; in other words, it is similar to a diagonal
matrix $D\in\mathbb{F}^{n\times n}$. Consider this $D$. Of course, the
diagonal entries of $D$ are the eigenvalues of $A$ (with multiplicities).
Since $A$ is similar to $D$, we have $\operatorname*{rank}
A=\operatorname*{rank}D$. But $D$ is diagonal; thus, its rank
$\operatorname*{rank}D$ equals the number of nonzero diagonal entries of $D$.
In other words, $\operatorname*{rank}D$ equals the number of nonzero
eigenvalues of $A$ (since the diagonal entries of $D$ are the eigenvalues of
$A$). In other words, $r$ equals the number of nonzero eigenvalues of $A$
(since $r=\operatorname*{rank}A=\operatorname*{rank}D$). In other words, the
matrix $A$ has exactly $r$ nonzero eigenvalues.
Label the eigenvalues of $A$ as $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$
(with multiplicities) in such a way that the first $r$ eigenvalues
$\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ are nonzero, while the remaining
$n-r$ eigenvalues $\lambda_{r+1},\lambda_{r+2},\ldots,\lambda_{n}$ are zero.
(This is clearly possible, since $A$ has exactly $r$ nonzero eigenvalues.)
Thus, $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ are exactly the nonzero
eigenvalues of $A$.
Lemma 3 yields
\begin{align*}
\det\left(  tI_n +A\right)   &  =\left(  t+\lambda_{1}\right)  \left(
t+\lambda_{2}\right)  \cdots\left(  t+\lambda_{n}\right)  =\prod_{i=1}
^{n}\left(  t+\lambda_{i}\right)  \\
&  =\left(  \prod_{i=1}^{r}\left(  t+\lambda_{i}\right)  \right)  \cdot\left(
\prod_{i=r+1}^{n}\left(  t+\underbrace{\lambda_{i}}
_{\substack{=0\\\text{(since }\lambda_{r+1},\lambda_{r+2},\ldots,\lambda
_{n}\text{ are zero)}}}\right)  \right)  \\
&  =\left(  \prod_{i=1}^{r}\left(  t+\lambda_{i}\right)  \right)
\cdot\underbrace{\left(  \prod_{i=r+1}^{n}t\right)  }_{=t^{n-r}}=\left(
\prod_{i=1}^{r}\left(  t+\lambda_{i}\right)  \right)  \cdot t^{n-r}.
\end{align*}
Now, Corollary 2 yields
\begin{align*}
&  \left(  \text{the sum of all principal }r\times r\text{-minors of
}A\right)  \\
&  =\left(  \text{the coefficient of }t^{n-r}\text{ in the polynomial
}\underbrace{\det\left(  tI_n +A\right)  }_{=\left(  \prod_{i=1}^{r}\left(
t+\lambda_{i}\right)  \right)  \cdot t^{n-r}}\right)  \\
&  =\left(  \text{the coefficient of }t^{n-r}\text{ in the polynomial }\left(
\prod_{i=1}^{r}\left(  t+\lambda_{i}\right)  \right)  \cdot t^{n-r}\right)  \\
&  =\left(  \text{the coefficient of }t^{0}\text{ in the polynomial }
\prod_{i=1}^{r}\left(  t+\lambda_{i}\right)  \right)  \\
&  \qquad\left(  \text{by Lemma 4, applied to }m=0\text{ and }k=n-r\text{ and
}p=\prod_{i=1}^{r}\left(  t+\lambda_{i}\right)  \right)  \\
&  =\left(  \text{the constant term of the polynomial }\prod_{i=1}^{r}\left(
t+\lambda_{i}\right)  \right)  \\
&  =\prod_{i=1}^{r}\lambda_{i}=\lambda_{1}\lambda_{2}\cdots\lambda_{r}\\
&  =\left(  \text{the product of all nonzero eigenvalues of }A\right)
\end{align*}
(since $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ are exactly the nonzero
eigenvalues of $A$). This proves Theorem 5. $\blacksquare$
Note that in the above proof of Theorem 5,
the diagonalizability of $A$ was used only to guarantee that $A$
has exactly $r$ nonzero eigenvalues and that all $n$ eigenvalues of $A$
belong to $\mathbb{F}$.
