51

The formula you're looking for can be obtained by differentiating Jacobi's formula
$$ \frac{\mathrm{d}}{\mathrm{d}t} \det A(t) = \det A(t) \cdot \operatorname{tr}\left( A^{-1} \frac{\mathrm{d}A}{\mathrm{d}t} \right) $$
with respect to a second parameter, say $s$:
\begin{multline}
\frac{\partial^2}{\partial s \partial t} \det A(s,t) =
\det A(s,t) \cdot \bigg[
...

33

At first, we prove that the determinant is non-zero, in other words, the matrix is non-singular. Assume the contrary, then by the linear dependency of the columns there exist real numbers $\lambda_1,\dots,\lambda_n$, not all equal to 0, such that $F(x_i):=\sum_j \frac{\lambda_j}{g(x_i-b_j)}=0$ for all $i=1,2,\dots,n$. But the equation $F(x)=0$ is a ...

answered Jul 8 '18 at 17:14

Fedor Petrov

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31

Here are some ideas how to decide the conjecture. (EDIT: In fact these ideas lead to a proof of the conjecture as Terry Tao explained in two comments below.)
As Christian Remling and Will Sawin showed, the conjecture is equivalent to $\det(I+T)\geq 0$ for any $T\in\mathrm{SO}^0(n,n)$.
We can assume that $-1$ is not an eigenvalue of $T$. Up to conjugacy, $...

30

I will work over a field of characteristic $0$ so that reductive algebraic groups are linearly reductive; presumably there is a way to eliminate this hypothesis. In this case, for an integer $n>1$ and for a divisor $m$ such that $n>m>1$, there does not exist $f$ as above. The point is to consider the critical locus of the determinant. The ...

30

I think a result of Hochster allows to get a quick proof that it is not possible to express the determinant of the generic $d \times d$ matrix as the determinant of a $k \times k$ matrices with entries being homogeneous forms of degree $\dfrac{d}{k}$, provided that $1 <k <d$.
I will work over an algebraically closed field.
Relying on some ...

30

I think the reference "Advanced Determinant Calculus" has a pointer to the answer. But I'll still elaborate for it is ingenious.
Suppose $x_i$'s and $y_j$'s, $1\leq i,j \leq N$, are numbers such that $x_i+y_j\neq 0$ for any $i,j$ combination, then the following identity (called Cauchy Alternant Identity) holds good:
$$
\det ~\left(\frac{1}{x_i+y_j}\right)_{...

26

The history of the Vandermonde notation is described, in the context of the Vandermonde determinant, in section 2.1 of A case of mathematical eponymy: the Vandermonde determinant (2010). It seems Lebesgue didn't like it because it could have induced a mix-up between indices and exponents, and that may be a reason it did not survive. Leibniz used a similar ...

answered Sep 28 '16 at 6:28

Carlo Beenakker

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25

To complement Fedor's answer, here is more explicit proof.
Let the original matrix be $G$. Let $D_x :=\text{Diag}(e^{x_1},\ldots,e^{x_n})$. Then, we can write
\begin{equation*}
G = D_x C D_b,\quad\text{where}\ C = \left[ \frac{1}{e^{2x_i}+e^{2b_j}}\right]_{i,j=1}^n.
\end{equation*}
To prove that $\det(G)>0$ it thus suffices to prove that $\det(C)>0$. ...

25

I won't discuss the permanent part of the question, but I think the other part can be done easily, even without Galois theory. Let $X = X(G)$ denote the character table of $G$ (rows indexed by complex irreducible characters of $G$, columns indexed by conjugacy classes of $G$). Then, by the orthogonality relations, $X\overline{X}^{T}$ is an integer diagonal ...

answered Jun 21 '20 at 9:16

Geoff Robinson

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Write the SVD of $A$, say $A=PDQ^T$ with $D$ diagonal with non-negative entries and $P\in O(n),Q\in O(m)$. Then $\sqrt{I_n + AA^T} = P\sqrt{1+D^2}P^T$
and $\sqrt{I_m+ A^TA} = Q\sqrt{1+D^2}Q^T$. This gives
$$
\begin{pmatrix}
\sqrt{I_n + AA^T} & A \\ A^T& \sqrt{I_m+A^TA}
\end{pmatrix}
=
\begin{pmatrix}
P & 0 \\
0 & Q
\end{pmatrix}
\begin{...

23

Let's look at $a_n=\det\left[\frac{(i-1)!}{(2j-1)!}\binom{i^2-\theta^2}j\right]_{i,j=1}^{n}$. By taking out common factors from rows we can write
$$a_n=\left(\prod_{i=1}^n (i-1)!(i^2-\theta^2)\right)\det\left[\frac{1}{(2j-1)!}\frac{1}{(i^2-\theta^2)}\binom{i^2-\theta^2}{j}\right]_{i,j=1}^{n}$$
$$=\left(\prod_{i=1}^n (i-1)!(i^2-\theta^2)\right)\det\left[\frac{...

answered May 11 '17 at 4:08

Gjergji Zaimi

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23

There is no such isomorphism (at least for $g \geq 9$).
In
O. Randal-Williams, The Picard group of the moduli space of r-Spin Riemann surfaces. Advances in Mathematics 231 (1) (2012) 482-515.
I computed the Picard groups of moduli spaces of Spin Riemann surfaces (for $g \geq 9$). Grothendieck--Riemann--Roch shows that, in the notation of that paper, the ...

answered Oct 10 '17 at 11:23

Oscar Randal-Williams

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22

Very nice problem!
Let me recall you that the determinant of $n \times n$ matrices with entries in $\{0,1\}$ is related to the one of $n+1 \times n+1$ matrices with entries in $\{-1,+1\}$: replace the zeros by $-1$'s and add a row of $-1$'s on the top and a column of ones on the right (you may want to read this arXiv).
I can now tell you that, besides ...

22

$\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\suml}{\sum\limits}
\newcommand{\sumS}{\suml_{S \in P}}
\newcommand{\prodl}{\prod\limits}
\newcommand{\prodS}{\prodl_{S \in P}}
\newcommand{\subs}{\subseteq}...

21

Let me add some more. If $A, B, C$ are positive semidefinite, then $$\det (A+B+C)+\det C\ge \det (A+C)+\det (B+C). \quad (\star)$$
When $C=0$, this reduces to OP's question.
A remarkable extension of ($\star$) were recently obtained by V. Paksoy, R. Turkmen, F. Zhang [ Electron. J. Linear Algebra 27 (2014)
332-341], which says that the determinant ...

21

Three vectors $v_1,v_2,v_3$ lie in a hyperplane $H:\alpha x+\beta y+\gamma z+\delta t=0$, in this plane we have $Q(v,v):=(\alpha x+\beta y)^2-(\gamma z+\delta t)^2=0,\forall v\in H$. Thus by polarization $Q(v,w)=\frac 14 (Q(v+w,v+w)-Q(v-w,v-w))=0$ for all $v,w\in H$ that yields a relation between columns of your matrix: if $v=(a,b,c,d), w=(A,B,C,D)$, then $Q(...

answered Oct 27 '19 at 6:45

Fedor Petrov

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21

We have $Af(A^TA)=f(AA^T)A$ for any reasonable function $f$, including $f(x)=\sqrt{1+x}$. This suffices to check for $f(x)=x^k$ when it is obvious, then approximate your function by a polynomial.

answered Jun 28 '21 at 10:53

Fedor Petrov

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20

The case $k=n$ is a consequence of the identity
$$\int \det(f_j(s_k))\det(g_j(s_k))\prod_{j=1}^N d\mu(s_j) = N!\ \det\left(\int d\mu(t) f_j(t)g_k(t)\right)$$
which I have seen under the names "Andreief identity" and also "Gram identity". The proof is elementary using the Leibniz formula for the determinant.

20

This question is too hard, because easier questions are already known to be hard.
The middle column is A046747 in the OEIS, which is essentially equivalent to A057982, the number of singular {±1}-valued matrices. The asymptotic behavior of this number was a longstanding open problem that was just recently solved by Tikhomirov.
As you mentioned yourself, ...

answered Jan 15 '19 at 22:24

Timothy Chow

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19

Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
...

answered Feb 10 '19 at 19:12

Fedor Petrov

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19

Here is a combinatorial proof. As my running example, I'll take $a=2$ and $n=3$, so I want to show
$$\det \begin{bmatrix} 1 & 3 & 1 \\ 1 & 6 & 5 \\ 1 & 10 & 15 \\ \end{bmatrix} = \det \begin{bmatrix} 5 & 14 \\ 14 & 42 \end{bmatrix}.$$
Remember the Lindström–Gessel–Viennot lemma -- Let $G$ be a directed acyclic graph with ...

answered Mar 1 '19 at 4:59

David E Speyer

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18

By Andreieff's identity:
$$
{\rm det}(A)=\frac{1}{n!}\int_{(0,\infty)^{n}}
{\rm det}[e^{-\frac{t_k}{2}}t_k^{\lambda_i-\frac{1}{2}}]_{1\le i,k\le n}\ \times\
{\rm det}[e^{-\frac{t_k}{2}}t_k^{\mu_j-\frac{1}{2}}]_{1\le k,j\le n}\ \ dt_1\cdots dt_n
$$
$$
=\frac{1}{n!}\int_{(0,\infty)^{n}}
\left(\prod_{l=1}^{n} \frac{e^{-t_l}}{t_l}\right)\times
{\rm det}[t_k^{\...

answered Jul 19 '18 at 15:34

Abdelmalek Abdesselam

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17

This is true, and in fact you can show a slightly more general fact:
$$\det_{1\le i,j\le n}\left( \binom{x_i}{2j}+ \binom{-x_i}{2j}\right)=\prod_{i=1}^n x_i^2 \prod_{i<j} (x_j^2-x_i^2) \prod_{j=1}^n \frac{2}{(2j)!}.$$
It is easy to show that this evaluates to $1$ when you set $x_i=i$. To prove this, notice that every column is an even polynomial in the $...

answered Jan 13 '14 at 0:50

Gjergji Zaimi

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16

Not all has been said about this question that is worth saying -- at
the very least, someone could have written down the version without the
absolute values; but more importantly, there are various other equally good proofs.
Notations and statement
Let me first state the result with proper signs and no absolute values.
Standing assumptions. The following ...

answered Aug 4 '16 at 15:47

darij grinberg

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16

We can just manipulate $C$ in the usual way by row operations: Subtract the last "row" from all the other "rows" (this is really several traditional row operations done at once). This produces
$$
\begin{pmatrix} A- B &0& 0 & \ldots & 0 &B-A \\
0 & A-B &0 &\ldots & 0 & B-A\\
&& \ldots &&&\\
B & B ...

answered Aug 25 '14 at 19:18

Christian Remling

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16

The answer is YES in every dimension, up to a sign. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore
$${\rm Pf}(\tilde A)^2=\det(I+iA)\det(I-iA).$$
On the other hand
$$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$
yields
$${\rm Pf}(\tilde A)^2=(\det(I+iA))^2.$$
Let us ...

answered Apr 19 '16 at 10:10

Denis Serre

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16

Your guess is correct. If the elements outside the diagonal have absolute values less than $1/(n-1)$, the matrix has 'diagonal dominance', thus it is nonsingular.
To make the answer self-contained, I give a proof. If $x=(x_1,\dots,x_n)^t$ satisfies $Ax=0$, take $k$ such that $|x_k|$ is maximal and look at $\sum a_{ki}x_i$. The summand $a_{kk}x_k$ has ...

15

let me assume $A$ is invertible, then you ask when
$$\det(1+X)=1+\det X,\;\;X=A^{-1}B $$
so if $X$ has eigenvalues $x_i$, $i=1,2,\ldots n$, you would need
$$\prod_{i}(1+x_i)=1+\prod_i x_i$$
basically you can take arbitrary values for $x_1,x_2,\ldots x_{n-1}$ and then the only requirement is that
$$x_n=\frac{1-U}{U-V},\;\;U=\prod_{i=1}^{n-1}(1+x_i),\;\;V=\...

answered Jul 8 '17 at 12:01

Carlo Beenakker

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15

Darij already gave a great elementary detailed exposition. I wanted to remark that all such kinds of results (including the postscript) are special cases of a classical result that (at least in the generality of semilattices) goes back to Lindström in Determinants on semilattices. For this question we are in the special case of the Boolean lattice.

answered Dec 7 '18 at 11:03

Gjergji Zaimi

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14

Update 1: Thanks to jjcale for pointing out a fatal flaw. Indeed $SO(n,n)$ has two components, see here, and it looks suspiciously like my $T$ below is in the wrong component. I don't really know what Wikipedia means by "preserving/reversing orientation," but certainly $T=\textrm{diag}(-1,1,-1,1)$ is in the wrong component, and my $T$ below feels like it ...

answered May 1 '15 at 17:30

Christian Remling

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