Determinant of $H_n^\alpha$ Let $H_n$ be a matrix of order $n$ such that its diagonal elements are all $\alpha$ and has $-1$ in the non diagonal entries.
Show that $det(H_n)=(\alpha +1 -n)(\alpha + 1)^n$.
I am baffled by this exercise. I tried assuming the result and expressing the determinant of $H_{n+1}$ in terms of the determinant of $H_n$, but I get to the formula $det(H_{n+1})=det(H_{n})(\alpha + 1)-(\alpha + 1)^n$ and I see no way of deducing where the formula comes from.
 A: Here is a direct proof using the powerful Matrix-Determinant lemma.
Let $1$ be the column vector with all entries equal to 1. We can write:
$$H_n+11^T=(\alpha+1)I_n.$$
Thus $$H_n=\beta I_n-11^T \ \ \ \ \text{with} \ \ \ \ \beta:=(\alpha+1).$$
Then use Matrix-Determinant lemma (https://en.wikipedia.org/wiki/Matrix_determinant_lemma) with $u=-1^T$ and $v=1$:
$$det(H_n)=\underbrace{(1-1^T(\beta I_n)^{-1} 1)}_{1-\dfrac{n}{\beta}}\underbrace{det(\beta I_n)}_{\beta^n}=(\beta-n)\beta^{n-1}=(\alpha+1-n)(\alpha+1)^{n-1}.$$
A: I think you mean $(\alpha+1-n)(\alpha+1)^{n-1}$
This is a real symmetric matrix, so has a basis of real eigenvectors.
The determinant is the product of the eigenvalues. You should be able
to guess what these are, and confirm they are eigenvalues.
A: This proof yields the desired result and the eigendecomposition of $H_n$.
Let $e$ denote the all-ones vector and $e_i$ denote the $i^\text{th}$ canonical basis vector of $\mathbb{C}^n$. Since every row-sum of $H_n$ is constant and equal to $\alpha + 1 -n$, it follows that $$H_ne = (\alpha + 1 -n)e,$$ 
i.e., $\alpha + 1 -n \in \sigma(H_n)$. 
For $i=2,\dots,n$, notice that
$$ H_n(e_i - e_1) = 
\begin{bmatrix} 
-1 \\
-1 \\
\vdots \\
-1 \\
\alpha \\
-1 \\
\vdots \\
-1
\end{bmatrix}-
\begin{bmatrix} 
\alpha \\
-1 \\
\vdots \\
-1 \\
-1 \\
-1 \\
\vdots \\
-1
\end{bmatrix} =
\begin{bmatrix} 
-(1+\alpha) \\
0 \\
\vdots \\
0 \\
\alpha+1 \\
0 \\
\vdots \\
0
\end{bmatrix}=
(\alpha+1)(e_i-e_1),$$
i.e., $\alpha+1 \in \sigma(H_n)$.
If 
$$ S_n := 
\begin{bmatrix} 
e \mid e_2-e_1\mid\cdots\mid e_n-e_1
\end{bmatrix} = 
\begin{bmatrix}
1 & -1 & \cdots & -1 \\
1 & 1 &  &  \\
\vdots &  & \ddots &  \\
1 & & & 1
\end{bmatrix},$$
then $\det(S_n)\neq 0$ (clearly, $S_n \sim I_n$) and $H_n = S_n D_n S_n^{-1}$, where
$$ D_n := 
\begin{bmatrix}
\alpha+1-n & &  \\
& \alpha+1 & &  \\
& & \ddots & & \\
& & & \alpha+1 
\end{bmatrix} \in M_n(\mathbb{C}).$$
Thus, $\det(H_n)=(\alpha+1-n)(\alpha+1)^{n-1}$.   
