It has been a while with linear algebra. Can someone give a hint to find the determinant of a $n\times n $ matrix where the non-diagonal elements are $1$:
$$
M=
\
\begin{bmatrix}
-\lambda & 1 & .. & 1 \\
1 & -\lambda & .. & 1 \\
.. & .. &.. & ..\\
1 & 1 & 1 & -\lambda
\end{bmatrix}
\
$$
There is a Leibniz formula for the det of $n\times n$ matrices which is:
$$\det(A) = \sum_{\sigma \in S_n} \left( \text{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}\right)$$
Applying this, I can see that:
$$\det(M) = \text{sgn}(fp_0)n_{fp_0} +\text{sgn}(fp_1)n_{fp1} \times (-\lambda) +…+\text{sgn}(fp_{n-1}) n_{fp_{n-1}}\times (-\lambda)^{n-1} $$
where $n_{fp_i}$ is the number of permutations with $i$ fixed points and $\text{sgn}(fp_i)$ is the sign of all such permutations. Is this logic going to lead to the answer? Thanks.
Best Answer
Hint: the matrix $$\begin{bmatrix}1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1\end{bmatrix}$$ is diagonalizable, with eigenvalue $0$ of multiplicity $n-1$, and eigenvalue $n$ of multiplicity $1$. So $$\det\begin{bmatrix}1-\lambda & 1 & \cdots & 1 \\ 1 & 1-\lambda & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1-\lambda\end{bmatrix} = (-\lambda)^{n-1}(n-\lambda).$$