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?
 A: 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, $\det(J+I) = \prod_{i=1}^n (\lambda_i + 1) = n+1$. $\square$
A: Here's a geometric proof: whether it's simpler/more intuitive than the row manipulations proof is a matter of taste.
Let $e_i$ be the elementary vector with a $1$ in the $i$th entry and $0$ everywhere else, and let $u$ denote the all-ones vector. We seek to find the determinant of the matrix $M$ whose $i$th row is $v_i:=u+e_i$. Since $M$ scales volumes by $\det M$, this determinant is the ratio
$$\det M=\frac{\operatorname{Volume}(\text{simplex with vertices }\{0,v_1,\dots,v_n\})}{\operatorname{Volume}(\text{simplex with vertices }\{0,e_1,\dots,e_n\})},$$
since $Me_i=v_i$. For simplex volumes, we use the formula $\frac1n\text{base}\cdot\text{height}$, with bases of each determined by $\{e_1,\dots,e_n\}$ and $\{v_1,\dots,v_n\}$. Since the $(n-1)$-simplex with vertices $\{v_1,\dots,v_n\}$ is just a translation of the simplex with vertices $\{e_1,\dots,e_n\}$, their bases have the same volume, and so $\det M$ is the ratio of the distances from $0$ to the planes determined by $\{v_1,\dots,v_n\}$ and $\{e_1,\dots,e_n\}$.
These planes are preserved under any permutation of the axes, and so the projection of $0$ down to each of these planes must be preserved by such transformations as well, and must be some scalar multiple of the all-ones vector $u$. The multiple of $u$ on the plane spanned by $\{e_1,\dots,e_n\}$ is $\frac1nu$, while the multiple of $u$ on the plane spanned by $\{v_1,\dots,v_n\}$ is $(1+\frac 1n)u$. So,
$$\det M=\frac{\left(1+\frac 1n\right)\|u\|}{\frac 1n\|u\|}=n+1.$$
A: 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
$$
p(-1)=\det(J_n+I_n)=1^{n-1}(n+1)=n+1
$$
