6
$\begingroup$

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?


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.

$\endgroup$
  • $\begingroup$ Did you try induction? $\endgroup$ – zarathustra Aug 14 '14 at 17:08
  • $\begingroup$ Induction combined with this should do it. $\endgroup$ – Daniel R Aug 14 '14 at 17:09
  • $\begingroup$ Induction is the way to go, but first, subtract the first row from all the other rows to get a much easier calculation. $\endgroup$ – fixedp Aug 14 '14 at 17:10
  • $\begingroup$ When $n=2$ seems to fail. $1-t^2$ RHS, $-2 t^3-3 t^2+1$ LHS, right? $\endgroup$ – carlosayam Aug 14 '14 at 17:14
  • 1
    $\begingroup$ The matrix size should be $n+1$; @caya gives the $n=2$ case. $\endgroup$ – Semiclassical Aug 14 '14 at 17:17
11
$\begingroup$

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

$\endgroup$
  • 5
    $\begingroup$ For clarity, first he is doing "Row N = Row N - Last Row" then he is doing "Last Row = Last Row - Row N" $\endgroup$ – DanielV Aug 14 '14 at 17:46
  • $\begingroup$ In the second step of the above comment, it makes sense to exchange references to "row" with "column". For further clarification to the answer, the sign of (n-1)t-1 is always positive (-1^2n); the cofactor expansion along the last row has only one term with a nonvanishing determinant ((n-1)t-1 itself) because the others all yield minors with a zero-column somewhere, and the determinant in the remaining term is simply an (n-1)x(n-1) diagonal with (-t - 1) as each element. That determinant yields (-t - 1)^(n-1), which here is shown with (-1)^(n-1) factored out of it. $\endgroup$ – Darren Ringer Feb 9 '17 at 20:04
7
$\begingroup$

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 $(t+1)$: $$ \det(t C - (t+1)I) = (-1)^n(t+1)^n \det\left(I - \frac{t}{t+1} \mathbf{1}\right) = (-1)^n(t+1)^n \left(1 - \frac{nt}{t+1}\right) $$

$\endgroup$
  • $\begingroup$ That's brilliant, nothing could be more elegant. :) $\endgroup$ – MGA Aug 14 '14 at 17:37
  • $\begingroup$ +1, but what do you mean by $cr$ and $rc$? Are these row and column vectors, abbreviated $r$ and $c$, respectively? (My assumption) $\endgroup$ – apnorton Aug 15 '14 at 3:44
  • $\begingroup$ Correct, they are row and column vectors. I stole that from the Wiki page on Determinant $\endgroup$ – Victor Liu Aug 15 '14 at 21:42
3
$\begingroup$

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 $I$).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.