Determinant of a matrix with $t$ in all off-diagonal entries.

Question

It seems from playing around with small values of $n$ that

$$ \det \left( \begin{array}{ccccc} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{array}\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1) $$

where $n$ is the size of the matrix.

How would one approach deriving (or at least proving) this formally?

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Motivation

This came up when someone asked what is the general solution to:

$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b},$$

and for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\implies a+b+c=1$ or $t=\frac{1}{2}\implies a=b=c$.

So I wanted to ensure that these are also the only solutions for the case with more variables.

Answer 1

Using elementary operations instead of induction is key. $$\begin{align} &\begin{vmatrix} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= \begin{vmatrix} -t-1 & 0 & 0 & \dots & t+1\\ 0 & -t-1 & 0 & \dots & t+1\\ 0 & 0 & -t-1 & \dots & t+1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= \begin{vmatrix} -t-1 & 0 & 0 & \dots & 0\\ 0 & -t-1 & 0 & \dots & 0\\ 0 & 0 & -t-1 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& (n - 1)t -1 \end{vmatrix}\\ &= (-1)^{n - 1}(t + 1)^{n - 1}((n - 1)t - 1) \end{align}$$

Answer 2

You can write the expression as $$ \det(t C - (t+1)I)$$ where $C = \mathbf{1}\mathbf{1}^T$ is the matrix of all $1$'s, formed by the column of ones times its transpose. Using the identity $\det(I+cr) = 1+rc$, you can first factor out $(t+1)$: $$ \det(t C - (t+1)I) = (-1)^n(t+1)^n \det\left(I - \frac{t}{t+1} \mathbf{1}\right) = (-1)^n(t+1)^n \left(1 - \frac{nt}{t+1}\right) $$

Answer 3

Note that your matrix is the sum of the matrix $T$ with all entries equal to $t$ and the matrix $-(1+t)I$. Therefore the determinant you are asking about is the value at $X=-(1+t)$ of the characteristic polynomial $\chi_{-T}$ of $-T$.

Since $T$ has rank (at most) $1$, its eigenspace for eigenvalue $0$ has dimension $n-1$, so the characteristic polynomial of $-T$ is $\chi_{-T}=X^{n-1}(X+nt)$ (the final factor must be that because the coefficient of $X^{n-1}$ in $\chi_{-T}$ is $\def\tr{\operatorname{tr}}-\tr(-T)=nt$). Now setting $X=-(1+t)$ gives $$ \det(-(1+t)I-T)=(-1-t)^{n-1}(-1-t+nt)=(-1-t)^{n-1}(-1+(n-1)t) $$ as desired.

This kind of question is recurrent on this site; see for instance Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$) and How to calculate the following determinants (all ones, minus $I$).

Answer 4

Here is another characterization of the result. For convenience, I will take the matrix size as $n+1$ rather than $n$. First, note that we may factor $t$ from each of the $n+1$ rows, and so the determinant may be written as $$ \det{[t(M-t^{-1} I_{n+1})]} =t^{n+1} \det(M-t^{-1} I_{n+1}) $$ where $(M)_{ij}=1-\delta_{ij}$ for $1\leq i,j\leq n+1$.

Next, observe that $M$ has $n$ independent eigenvectors of the form $\hat{e}_i-\hat{e}_{n+1}$ ($1\leq i\leq n$), each with eigenvalue $-1$; in addition, $M$ also has the eigenvector $\sum_{i=1}^{n+1}\hat{e}_i$ with eigenvalue $n$. Consequently the characteristic polynomial of $M$ in powers of $t^{-1}$ is $$\det(M-t^{-1} I_{n+1})=(-1-t^{-1})^n (n-t^{-1}) = t^{-n-1}\cdot (-1)^n (1+t)^n (nt+1)$$ and so the prior result yields the desired identity.