4

To count the number of spanning trees of a complete graph of order $n$ one can use Kirchhoff matrix theorem and arrive at the exact answer $n^{n-2}$. But in doing so, one should know how to evaluate the determinant of this $n-1\times n-1$ matrix? \begin{bmatrix} n-1 & -1 & \cdots & -1 \\ -1& n-1 & \cdots & -1\\ \vdots&\vdots &\ddots &\vdots \\ -1& -1 &\cdots & n-1 \end{bmatrix} I'm asking an efficient method to calculate the determinant of this matrix.

albo
  • 1,034

1 Answers1

9

$$B=\begin{bmatrix} n-1 & -1 & \cdots & -1 \\ -1& n-1 & \cdots & -1\\ \vdots&\vdots &\ddots &\vdots \\ -1& -1 &\cdots & n-1 \end{bmatrix}=nI_{n-1}-\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}\cdot\begin{bmatrix}1&1&\cdots&1\end{bmatrix}=nI_{n-1}-A$$

Now, $\text{rk}A=1$ and $$A\cdot\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}=\begin{bmatrix}n-1\\n-1\\\vdots\\n-1\end{bmatrix}$$

Hence, $A$ can be diagonalized

$P^{-1}AP=\text{diag}(n-1,0,\ldots,0)$.

Hence $\det(B)=\det(P^{-1}BP)=\det(nI_{n-1}-P^{-1}AP)=\det(\text{diag}(1,n,\ldots,n))=n^{n-2}$

  • Here "A can be diagonalized" is due to the algebraic and geometric multiplicities are same, then $P^{-1}$ can be valid. (Detailed reasons see this). Otherwise, the geometric multiplicity $\gamma _{A}(\lambda )$ may be smaller than algebraic multiplicity $\mu _{A}(\lambda )$ (one example). (Notice algebraic multiplicity $\ge$ geometric multiplicity. This is shown here) – An5Drama Feb 01 '24 at 10:03