Rank of the $n \times n$ matrix with ones on the main diagonal and $a$ off the main diagonal

Let

$$\mathrm M_n (a) := \begin{bmatrix} 1 & a & a & \dots & a & a\\ a & 1 & a & \dots & a & a\\ a & a & 1 & \dots & a & a\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a & a & a & \dots & 1 & a\\ a & a & a & \dots & a & 1\end{bmatrix} = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$$

The eigenvalues of rank-$1$ matrix $a 1_n 1_n^{\top}$ are

• $\color{blue}{0}$ with multiplicity $n-1$.
• $\color{blue}{n a}$ with multiplicity $1$.

Thus, the eigenvalues of $\mathrm M_n (a) = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$ are

• $\color{blue}{1-a}$ with multiplicity $n-1$.
• $(1-a) + na = \color{blue}{(n-1) \, a + 1}$ with multiplicity $1$.

We could also have arrived at this conclusion computing the characteristic polynomial of $\mathrm M_n (a)$

$$\begin{array}{rl} \det ( s \mathrm I_n - \mathrm M_n (a) ) &= \det \left( (s-(1-a)) \mathrm I_n - a 1_n 1_n^{\top} \right)\\ &= \det \left( (s-(1-a)) \left( \mathrm I_n - \frac{a}{s-(1-a)} 1_n 1_n^{\top} \right) \right)\\ &= (s-(1-a))^n \cdot \det \left( \mathrm I_n - \frac{a}{s-(1-a)} 1_n 1_n^{\top} \right)\\ &= (s-(1-a))^n \cdot \left(1 - \frac{n a}{s-(1-a)}\right)\\ &= (s-(1-a))^{n-1} \cdot \left(s-(1-a) - n a\right)\\ &= (s-(1-a))^{n-1} \cdot \left( s - ((n-1) \, a + 1) \right)\end{array}$$

where the matrix determinant lemma was used. If

$$a \in \left\{ -\frac{1}{n-1}, 1 \right\}$$

then $\mathrm M_n (a)$ is singular. Thus, using the rank-nullity theorem, we conclude that

$$\boxed{\mbox{rank} (\mathrm M_n (a)) = \begin{cases} 1 & \text{if } a = 1\\ n-1 & \text{if } a = -\frac{1}{n-1}\\ n & \text{otherwise}\end{cases}}$$

If you haven't figured it out yet here's the solution: if $a=1$ $\mathrm{Rank}(A)=1$ otherwise $\mathrm{Rank}(A)=n$ where $$A:= \begin{pmatrix} 1&a&a&\cdots&a\\ a&1&a&\cdots&a\\ a&a&1&\cdots&a\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a&a&1&\cdots&1 \end{pmatrix}$$

You've already shown that you know if $a\in\{0,1\}$. So as for the rest, your close to a solution. The next set of Row equations are as follows $R_i-R_1 \to R_i$ such that $1<i\leq n$. This gives the matrix,

$$A_2= \begin{pmatrix} 1&a&a&\cdots&a\\ -1&1&0&\cdots&0\\ -1&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&\cdots&1 \end{pmatrix}$$

The next Row equation is $\frac{R_1-\sum_{i=2}^n aR_i}{1+a(n-1)}\to R_1$. This gives the matrix

$$A_3= \begin{pmatrix} 1&0&0&\cdots&0\\ -1&1&0&\cdots&0\\ -1&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&\cdots&1 \end{pmatrix}$$

The next Row equations are $R_i+R_1\to R_i$ where $1<i\leq n$. This gives $$A_4= \begin{pmatrix} 1&0&0&\cdots&0\\ 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1 \end{pmatrix}=I$$

Note that $\mathrm{Rank}(A)=\mathrm{Rank}(I)=n$ as desired.

Karma made note that if $a=\frac{1}{1-n}$ then $$A_3= \begin{pmatrix} 0&0&0&\cdots&0\\ -1&1&0&\cdots&0\\ -1&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&\cdots&1 \end{pmatrix}$$

This implies that for $a=\frac{1}{1-n}$ the $\mathrm{Rank}(A)=n-1$.