Is Sum of Principal Minors Equals to Pseudo Determinant?

Question

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?

Answer 1

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 $\mathbb{F}$.