<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] I'm going to elaborate on Exodd's comment. I'm inclined to agree that this is the simplest proof, provided you're comfortable with the basic constructions of linear algebra.
Lemma: If $A \in \mathbf{C}^{n \times n}$ has eigenvalues (counting multiplicity) $(\lambda_i)_{i=1}^n$, then $A + I$ has eigenvalues $(\lambda_i + 1)_{i=1}^n$.
Proof: Observe that $\det(xI - (A + I)) = \det((x-1)I - A)$. $\square$

Proposition: If $J \in \mathbf{C}^{n \times n}$ is the all-ones matrix, then $\det(I+J) = n+1$.
Proof: Since $J$ is rank 1, hence it has exactly one nonzero eigenvalue. Since $\text{trace}(J) = \sum_{i=1}^n \lambda_i$, it follows that the unique nonzero eigenvalue $\lambda_1 = n$. By the lemma, $\det(J+I) = \prod_{i=1}^n (\lambda_i + 1) = n+1$. $\square$
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