Prove determinant of $n \times n$ matrix is $(a+(n-1)b)(a-b)^{n-1}$? Prove $\det(A)$ is $(a+(n-1)b)(a-b)^{n-1}$ where $A$ is $n \times n$ matrix with $a$'s on diagonal and all other elements $b$, off diagonal.
 A: Note that for $\lambda=a-b$ we have
$$
A-\lambda I
=
\begin{bmatrix}
b      & \cdots & b \\
\vdots & \ddots & \vdots \\
b      & \cdots & b
\end{bmatrix}\tag{1}
$$
The matrix in $(1)$ has rank $1$ so its nullspace has dimension $n-1$. Hence $\lambda_1=a-b$ is an eigenvalue of $A$ whose geometric multiplicity is $n-1$.
Now, note that for $\lambda=a+(n-1)b$ we have
$$
A-\lambda I=
\begin{bmatrix}
(1-n)b & \cdots & b \\
\vdots & \ddots & \vdots \\
b      & \cdots & (1-n)b
\end{bmatrix}\tag{2}
$$
The columns of the matrix in $(2)$ sum to $0$ so it has a nontrivial nullspace.
That is, $\lambda_2=a-(n-1)b$ is an eigenvalue for $A$. Because $\lambda_1$ has geometric multiplicity $n-1$, the geometric multiplicity of $\lambda_2$ must be $1$. Hence the characteristic polynomial of $A$ is
$$
\det(A-\lambda I)=(\lambda_1-\lambda)^{n-1}(\lambda_2-\lambda)\tag{3}
$$
Finally, plugging in $\lambda=0$ into $(3)$ gives the result.
A: For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal but not orthonormal) eigenvectors for the $n$ by $n$ case can be read from the columns of
$$    
 \left(  \begin{array}{rrrrrrrrrr}
  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  2  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  3  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  4  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  5  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  6  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  7  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  8  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  9   
\end{array}
  \right).
  $$
Note: I made this up. My own self.
A: Let 
$A=\begin{bmatrix}
a & b & b & b & \ldots & b \\
b & a & b & b & \ldots & b \\
b & b & a & b & \ldots & b \\
b & b & b & a & \ldots & b \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & b & \ldots & a 
\end{bmatrix}$
a $n\times n$ matrix, cause adding/substracting a scalar multiple of a row to another row doesn't change the value of the determinant of a matrix we can add rows 2nd, 3rd, 4th, $\dots$ and $n$th to the first row, so
$$\det (A) = \det \begin{bmatrix}
a+(n-1)b & a+(n-1)b & a+(n-1)b & a+(n-1)b & \ldots & a+(n-1)b \\
b & a & b & b & \ldots & b \\
b & b & a & b & \ldots & b \\
b & b & b & a & \ldots & b \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & b & \ldots & a 
\end{bmatrix}$$
Now, we can factor out $a+(n-1)b$ from the first row
$$\det (A) = \left[a+(n-1)b\right]\det \begin{bmatrix}
1 & 1 & 1 & 1 & \ldots & 1 \\
b & a & b & b & \ldots & b \\
b & b & a & b & \ldots & b \\
b & b & b & a & \ldots & b \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & b & \ldots & a 
\end{bmatrix}$$
Also, if we substract $b$ times row 1 to rows 2nd, 3rd, 4th, $\ldots$ and $n$th we get
$$\det (A) = \left[a+(n-1)b\right]\det \begin{bmatrix}
1 & 1 & 1 & 1 & \ldots & 1 \\
0 & a-b & 0 & 0 & \ldots & 0 \\
0 & 0 & a-b & 0 & \ldots & 0 \\
0 & 0 & 0 & a-b & \ldots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \ldots & a-b 
\end{bmatrix}$$
The last matrix is a lower triangular matrix and its determinant is the product of elements in the main diagonal. Therefore
$$\det (A) = \left[a+(n-1)b\right](a-b)^{n-1}$$
