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How do you attempt to solve the deterimnant of $$D_n=\begin{vmatrix} 0 & 1 & 1 & 1 & \cdots & 1 \\ 1 & 0 & x & x & \cdots & x \\ 1 & x & 0 & x & \cdots & x \\ 1 & x & x & 0 & \ddots & \vdots \\ \vdots & \vdots & \vdots & x & \ddots & x \\ 1 & x & x & \cdots & x & 0 \end{vmatrix}$$

My approach is to try and create a similar matrix with finite dimensions (something like a $4\times 4$) and then try and simplify it, but I always get stuck in the process of simplifying...

In general, how do you approach these types of questions?

Xander Henderson
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Tomergt45
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3 Answers3

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The first approach is to compute the first few values. Here they are (WA): $$ \begin{array}{c} n& 1 & 2 & 3 & 4 & 5 \\ D_n&0 & -1 & 2x & -3x^2 & 4x^3 \end{array} $$ The pattern is clear. Try a proof by induction.

lhf
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By Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)

we know that the determinant is $$ (-1)^{n-1}(n-1)x^{n} $$ for the matrix, where the one's are replaced by $x$. But multiplying the first row and column by $x$, we can obtain such a matrix from the given one. The determinant changes by a factor $x^2$ then. So the determinant of our matrix then is $$ (-1)^{n-1}(n-1)x^{n-2}. $$

Dietrich Burde
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We have $$x^2D_n = \begin{vmatrix} 0 & x & x & \cdots & x \\ x & 0 & x & \cdots & x \\ x & x & 0 & \cdots & x \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x & x & x & \cdots & 0 \end{vmatrix} = x^n\begin{vmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 0 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & 0 \end{vmatrix} = x^n\det(J-I)$$

where $J$ is the matrix with all entries equal to $1$. We know that $J$ has eigenvalues $0,0, \ldots, 0, n$ with multiplicity. Indeed, the respective basis of eigenvectors are $$e_2-e_1, e_3-e_2,\ldots, e_n-e_{n-1}, e_1+e_2+\cdots+e_n.$$The matrix $J-I$ therefore has eigenvalues $-1,-1, \ldots, -1, n-1$ with multiplicity. Determinant is the product of eigenvalues so $$\det(J-I)=(-1)^{n-1}(n-1)$$

We conclude $D_n = (-1)^{n-1}(n-1)x^{n-2}$.

mechanodroid
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