Vtgyjfy

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?

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 $A=left( a_{i,j}right)
  _{1leq ileq n, 1leq jleq m}$ be an $ntimes m$-matrix (over some ring).
  Let $U=left{ u_{1}<u_{2}<cdots<u_{p}right} $ be a subset of $left{
  1,2,ldots,nright} $, and let $V=left{ v_{1}<v_{2}<cdots<v_{q}right}
  $ be a subset of $left{ 1,2,ldots,mright} $. Then, $A_{U,V}$ shall
  denote the submatrix $left( a_{u_{i},v_{j}}right) _{1leq ileq p, 1leq
  jleq q}$ 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{ 1,3,4right} ,left{ 2,4right} }
  =
  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 $ntimes n$
  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 $ntimes
n$-matrix over $mathbb{K}$. Let $xinmathbb{K}$. Then,
begin{align}
detleft( A+xI_n right) & =sum_{Psubseteqleft{ 1,2,ldots
,nright} }detleft( A_{P,P}right) x^{n-leftvert Prightvert
}
label{darij.eq.t1.1}
tag{1}
\
& =sum_{k=0}^{n}left( sum_{substack{Psubseteqleft{ 1,2,ldots
,nright} ;\leftvert Prightvert =n-k}}detleft( 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 $operatorname*{sub}nolimits_{wleft(
Pright) }^{wleft( Pright) }A$ instead of $A_{P,P}$). $blacksquare$

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

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

Lemma 3. Let $A$ be an $ntimes n$-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[ tright] $, we have
begin{equation}
detleft( tI_n +Aright) =left( t+lambda_{1}right) left(
t+lambda_{2}right) cdotsleft( t+lambda_{n}right) .
end{equation}

Proof of Lemma 3. The eigenvalues of $A$ are defined as the roots of the
characteristic polynomial $detleft( tI_n -Aright) $ of $A$. (You may be
used to defining the characteristic polynomial of $A$ as $detleft(
A-tI_n right) $ instead, but this makes no difference: The polynomials
$detleft( tI_n -Aright) $ and $detleft( A-tI_n right) $ differ
only by a factor of $left( -1right) ^{n}$ (in fact, we have $detleft(
A-tI_n right) =left( -1right) ^{n}detleft( tI_n -Aright) $), and
thus have the same roots.)

Also, the characteristic polynomial $detleft( tI_n -Aright) $ 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, $detleft( tI_n -Aright) $ is
a monic polynomial of degree $n$ and has roots $lambda_{1},lambda_{2}
,ldots,lambda_{n}$. Thus,
begin{equation}
detleft( tI_n -Aright) =left( t-lambda_{1}right) left(
t-lambda_{2}right) cdotsleft( 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) cdotsleft( t-lambda_{n}right) $). Substituting
$-t$ for $t$ in this equality, we obtain
begin{align*}
detleft( left( -tright) I_n -Aright) & =left( -t-lambda
_{1}right) left( -t-lambda_{2}right) cdotsleft( -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( -1right) ^{n}underbrace{prod_{i=1}^{n}left( t+lambda
_{i}right) }_{=left( t+lambda_{1}right) left( t+lambda_{2}right)
cdotsleft( t+lambda_{n}right) } \
& = left( -1right) ^{n}left(
t+lambda_{1}right) left( t+lambda_{2}right) cdotsleft(
t+lambda_{n}right) .
end{align*}
Comparing this with
begin{equation}
detleft( underbrace{left( -tright) I_n -A}_{=-left( tI_n
+Aright) }right) =detleft( -left( tI_n +Aright) right) =left(
-1right) ^{n}detleft( tI_n +Aright) ,
end{equation}
we obtain
begin{equation}
left( -1right) ^{n}detleft( tI_n +Aright) =left( -1right)
^{n}left( t+lambda_{1}right) left( t+lambda_{2}right) cdotsleft(
t+lambda_{n}right) .
end{equation}
We can divide both sides of this equality by $left( -1right) ^{n}$, and
thus obtain $detleft( tI_n +Aright) =left( t+lambda_{1}right)
left( t+lambda_{2}right) cdotsleft( 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 $pinmathbb{F}left[ tright] $ be a polynomial. Then,
begin{align*}
& left( text{the coefficient of }t^{m+k}text{ in the polynomial }pcdot
t^{k}right) \
& =left( text{the coefficient of }t^{m}text{ in the polynomial }pright)
.
end{align*}

Proof of Lemma 4. The coefficients of the polynomial $pcdot t^{k}$ 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 $ntimes n$-matrix over a field
$mathbb{F}$. Let $r=operatorname*{rank}A$. Then,
begin{align*}
& left( text{the product of all nonzero eigenvalues of }Aright) \
& =left( text{the sum of all principal }rtimes rtext{-minors of
}Aright) .
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}
Ainleft{ 0,1,ldots,nright} $ (since $A$ is an $ntimes n$-matrix).

The matrix $A$ is diagonalizable; in other words, it is similar to a diagonal
matrix $Dinmathbb{F}^{ntimes n}$. 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*}
detleft( tI_n +Aright) & =left( t+lambda_{1}right) left(
t+lambda_{2}right) cdotsleft( t+lambda_{n}right) =prod_{i=1}
^{n}left( t+lambda_{i}right) \
& =left( prod_{i=1}^{r}left( t+lambda_{i}right) right) cdotleft(
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)
cdotunderbrace{left( prod_{i=r+1}^{n}tright) }_{=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 }rtimes rtext{-minors of
}Aright) \
& =left( text{the coefficient of }t^{n-r}text{ in the polynomial
}underbrace{detleft( tI_n +Aright) }_{=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) \
& qquadleft( text{by Lemma 4, applied to }m=0text{ and }k=n-rtext{ 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}cdotslambda_{r}\
& =left( text{the product of all nonzero eigenvalues of }Aright)
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 $mathbb{F}$.