What explains this $n \times n$ determinant pattern?

Question

$\mathbf{SETUP}$

I've been studying a physics model that involves inverting an $n \times n$ matrix of this form:

$$ \mathbf{A}^{-1}_n(x)= \begin{pmatrix} 1 & -x & -x & -x & ... \\ -x & 1 & -x & -x & ... \\ -x & -x & 1 & -x & ... \\ -x & -x & -x & 1 & ... \\ ... & ... & ... & ... & ... \end{pmatrix} ^{-1} $$ which made me interested in when it is not invertible, i.e. $\det(\mathbf{A}_n(x)) = 0$.

I managed to derive a nice reason for why: $$ \det(\mathbf{A}_n(x=\tfrac{1}{n-1})) = 0 $$

which is because the determinant is equivalent to the polynomial:

$$ \det(\mathbf{A}_{n+1}) = (x+1)^n (1-nx) $$

(Try it! It works :) - Incidentally, would this be a common result in some textbook?)

$\mathbf{QUESTION}$

But then, I playfully decided to just make a single (non-diagonal) matrix entry equal to zero, keeping all the others still at $\tfrac{1}{n-1}$, and stumbled upon this other pattern:

$$ \det \begin{pmatrix} 1 & 0 & \tfrac{1}{1-n} & \tfrac{1}{1-n} & ... \\ \tfrac{1}{1-n} & 1 & \tfrac{1}{1-n} & \tfrac{1}{1-n} & ... \\ \tfrac{1}{1-n} & \tfrac{1}{1-n} & 1 & \tfrac{1}{1-n} & ... \\ \tfrac{1}{1-n} & \tfrac{1}{1-n} & \tfrac{1}{1-n} & 1 & ... \\ ... & ... & ... & ... & ... \end{pmatrix} = \frac{n^{n-2}}{(n-1)^n} $$

(The above works regardless of which non-diagonal entry you choose to make zero, of course.)

Can anyone help me understand why? Might this be related to MacMahon's Master Theorem? Or is there some more obvious way to think about this result?

Answer 1

Let $B$ be the $n\times n$ matrix (with $n>2$) whose entries are all ones except the entry at row $1$, column $2$ (as in the question), which is zero. Then it is easy to verify that $B^3=nB^2-B$ (see below). Since the eigenvalue $0$ of $B$ is obviously of multiplicity $n-2$ (it's easy to exhibit the eigenvectors), we get $$\det(\lambda I-B)=\lambda^{n-2}(\lambda^2-n\lambda+1).$$

The matrix in the question is $(nI-B)/(n-1)$, so we put $\lambda=n$ and get the expected result.

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To verify $B^3=nB^2-B$, introduce the following $n\times n$ matrices:

• $S$ — the "all ones" matrix: $S_{i,j}=1$;
• $R$ — the matrix with the first row of ones: $S_{i,j}=[i=1]$;
• $C$ — the matrix with the second column of ones: $S_{i,j}=[j=2]$;
• $U$ — the matrix with the only nonzero entry at $(1,2)$: $S_{i,j}=[i=1\land j=2]$.

Now we build a multiplication table: \begin{align} SS &= nS & SR &= S & SC &= nC & SU &= C \\ RS &= nR & RR &= R & RC &= nU & RU &= U \\ CS &= S & CR &= 0 & CC &= C & CU &= 0 \\ US &= R & UR &= 0 & UC &= U & UU &= 0 \end{align} Since $B=S-U$, we get \begin{align} B^2&=SS-US-SU+UU \\ &=nS-R-C; \\ B^3&=(nS-R-C)(S-U) \\ &=nSS-RS-CS-nSU+RU+CU \\ &=n^2S-nR-S-nC+U \\ &=n(nS-R-C)-(S-U) \\ &=nB^2-B. \end{align}

Answer 2

The matrix in question is $$ M =\left(1-\frac{1}{1-n}I_n\right)+\frac{1}{1-n}(ee^T-e_1e_2^T) =\frac{1}{n-1}\left[nI_n-(ee^T-e_1e_2^T)\right]. $$ Hence $\det M=\dfrac{f(n)}{(n-1)^n}$, where $f(x)$ is the characteristic polynomial of $$ ee^T-e_1e_2^T=\pmatrix{e&e_1}\pmatrix{e^T\\ -e_2^T}. $$ By Sylvester’s secular theorem^[remark], $f(x)=x^{n-2}g(x)$, where $g(x)$ is the characteristic polynomial of $$ \pmatrix{e^T\\ -e_2^T}\pmatrix{e&e_1}=\pmatrix{n&1\\ -1&0}. $$ Therefore $g(x)=(x-n)x+1$ and $$ \det M =\dfrac{f(n)}{(n-1)^n} =\dfrac{n^{n-2}g(n)}{(n-1)^n} =\dfrac{n^{n-2}}{(n-1)^n}\ . $$

Remark. Sylvester’s secular theorem: for any two matrices $A$ and $B$ such that $AB$ is $n\times n$ and $BA$ is $m\times m$ with $n\ge m$, the characteristic polynomial of $AB$ is $x^{n-m}$ times the characteristic polynomial of $BA$. Simple proofs can be found in this answer by M.H. and this answer by Bill Dubuque.

Answer 3

In user1551's answer, the matrix of interest was expressed using two rank-one updates to the identity matrix. In this answer I'll show how to perform these rank-one updates sequentially.

First, note that the original matrix is, up to a multiplicative factor, itself a rank-one update of the $n$-by-$n$ identity matrix $I_n$: $$A_n(x)=(1+x)I_n-x u u^\top$$ where $u$ is a matrix of all ones. The matrix determinant lemma yields the updated determinant

$$\det{A_n(x)} = [1+(-x u^\top)(1+x)^{-1} I_n u](1+x)^n =(1+x-n x)(1+x)^{n-1}$$ In particular $\det A_{n+1}(x)=(1-nx)(1+x)^n$, as indicated in the OP. We can perform another rank-one update on $A_n(x)$ to eliminate the offending matrix element, but first we will need the inverse of $A_{n}(x)$. For this, the Sherman-Morrison formula yields the inverse \begin{align} A_n(x)^{-1} &=(1+x)^{-1} I_n-(1+x)^{-1} I_n \frac{(-xu)u^\top}{1+(-x u^\top)(1+x)^{-1} I_n u}(1+x)^{-1} I_n\\ &=(1+x)^{-1}I_n +x(1+x)^{-2} \frac{u u^\top}{1-x n(1-x)^{-1}}\\ &=(1+x)^{-1}I_n +\frac{x(1+x)^{-1}}{1+x-n x}u u^\top \end{align} We now consider the rank-one update $A_n'(x):=A_n(x)+x e_j e_k^\top$ with $j\neq k$. Another use of the matrix determinant lemma yields

\begin{align} \det A_n'(x) &=[1+x e_2^\top (A_n(x)^{-1} )e_1]\det(A_n(x))\\ &=\left[1+x(1+x)^{-1} e_k^\top e_j +\frac{x^2(1+x)^{-1}}{1+x-nx}e_k^\top u u^\top e_j\right](1+x-n x)(1+x)^{n-1}\\ &=\left[(1+x-nx)(1+x)+x^2\right](1+x)^{n-2} \end{align} (I could have covered the case $j=k$ using a Kronecker delta, but I didn't feel like doing so for now.) Finally, when $x=1/(n-1)$ we obtain $$\det A_{n}'(1/(n-1))=\frac{1}{(n-1)^2}\left(\frac{n}{n-1}\right)^{n-2}=\frac{n^{n-2}}{(n-1)^n}$$ in agreement with the OP.