<s>[INST] Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
  1 & 0 & 0 &\cdots & 0 \\
  0 & 1 & 0 &\cdots & 0 \\
  0 & 0 & 1 &\cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & 1
 \end{pmatrix} +\begin{pmatrix}
  1 & 1 & 1 & \cdots & 1 \\
  1 & 1 & 1 & \cdots & 1 \\
  1 & 1 & 1 & \cdots & 1 \\
  \vdots &  \vdots & \vdots & \ddots & \vdots \\
  1 & 1 & 1 &  \cdots & 1
 \end{pmatrix}\right| =\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  1 & 2 & 1 & \cdots & 1 \\
  1 & 1 & 2 & \cdots & 1 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & 1 & 1 & \cdots & 2
 \end{matrix} \right|=n+1.$$
I can prove it by adding and subtracting rows and columns without changing the determinant:

*

*subtracting the first row from the others $
\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  1 & 2 & 1 & \cdots & 1 \\
  1 & 1 & 2 & \cdots & 1 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & 1 & 1 & \cdots & 2
 \end{matrix} \right|=
\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|$


*subtracting the other rows from the first $\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|=
\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|$


*adding the other columns to the first $\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|=
\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  0 & 1 & 0 & \cdots & 0 \\
  0 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & 1
 \end{matrix} \right|$


*and clearly $\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  0 & 1 & 0 & \cdots & 0 \\
  0 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & 1
 \end{matrix} \right| =n+1$
but I feel there should be something more intuitive.
What would be a more intuitive or insightful way to prove this?
 [/INST] The eigenvalues of $J_n$ are $0$ with geometric multiplicity $n-1$ (because the rank of $J_n$ is one) and $n$ with algebraic multiplicity at least $1$ (an eigenvector is the “all $1$” column vector).
Thus the algebraic multiplicity of $0$ is $n-1$ and of $n$ is $1$. Hence the characteristic polynomial of $J_n$ is
$$
p(x)=\det(J_n-xI_n)=(0-x)^{n-1}(n-x)
$$
For $x=-1$, we get
$$
p(-1)=\det(J_n+I_n)=1^{n-1}(n+1)=n+1
$$
</s>