Prove that matrix is invertible I have a symmetric real-valued matrix with the following structure:
\begin{bmatrix}
1 + (n-1)\alpha_{1}^2 & 1-\alpha_{1}\alpha_{2} & ... & 1-\alpha_{1}\alpha_{n} \\
1-\alpha_{1}\alpha_{2} & 1+(n-1)\alpha_{2}^2 & ... & 1-\alpha_{2}\alpha_{n} \\
\vdots & \vdots & \ddots & \vdots\\
1-\alpha_{1}\alpha_{n} & 1-\alpha_{2}\alpha_{n} & ... & 1+(n-1)\alpha_{n}^2
\end{bmatrix}
I am trying to show that this matrix has non-zero determinant for all $n$ and therefore is invertible. I am seeing this is the case empirically. I would appreciate any hints or ideas to prove this formally.
Thanks
Daniel
 A: Let us assume that $\alpha_i\neq 0$ for $i\in\{1,\ldots,n\}.$
Let $X$ be the original matrix and 
$$
A = n\cdot \begin{pmatrix}
\alpha_1^2 & & & \\
& \alpha_2^2 & & \\
& & \ddots & \\
& & & \alpha_n^2
\end{pmatrix}
$$
and
$$
U = \begin{pmatrix}
1 & \alpha_1 \\
1 & \alpha_2 \\
\vdots & \vdots \\
1 & \alpha_n
\end{pmatrix}
\;\;\; , \;\;\;
C = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\;\;\; , \;\;\;
V = \begin{pmatrix}
1 & 1 & \cdots & 1 \\
-\alpha_1 & -\alpha_2 & \cdots & -\alpha_n
\end{pmatrix}
$$
Then $X = A+UCV,$ $A$ is invertible and we can use the Woodbury matrix identity
$$
(A+UCV)^{-1} = A^{-1} - A^{-1}U(C^{-1}+VA^{-1}U)^{-1}VA^{-1}
$$
if $C^{-1}+VA^{-1}U$ is invertible.
After some calculations, we get
$$
C^{-1}+VA^{-1}U = I + 
\frac{1}{n}\begin{pmatrix}
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i^2} &
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} \\
-\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} & -n
\end{pmatrix}
=
\frac{1}{n}\begin{pmatrix}
n+\sum\limits_{i=1}^{n}\frac{1}{\alpha_i^2} &
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} \\
-\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} & 0
\end{pmatrix}
$$
which is obviously invertible if and only if
$$
\sum\limits_{i=1}^{n}\frac{1}{\alpha_i} \neq 0
$$
In case that $\sum\limits\frac{1}{\alpha_i} = 0,$ it can easily be seen that
$$
X\cdot\begin{pmatrix} \alpha_1^{-1} \\ \alpha_2^{-1} \\ \vdots \\ \alpha_n^{-1} 
\end{pmatrix} = 0
$$
which means that $X$ cannot be invertible in this case.
It can also easily be seen that the matrix $X$ is not invertible if two or more of the $\alpha_i$ are $0.$
So the only case left is the case in which there is exactly one $i$ such that $\alpha_i=0.$ Wolog $\alpha_1 = 0.$ We can modify the first part of the proof a little bit:
$$
A = n\cdot \begin{pmatrix}
1 & & & \\
& \alpha_2^2 & & \\
& & \ddots & \\
& & & \alpha_n^2
\end{pmatrix}
$$
and
$$
U = \begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & \alpha_2 \\
\vdots & \vdots & \vdots \\
0 & 1 & \alpha_n
\end{pmatrix}
\;\;\; , \;\;\;
C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\;\;\; , \;\;\;
V = \begin{pmatrix}
-n & 0 & \cdots & 0 \\
1 & 1 & \cdots & 1 \\
0 & -\alpha_2 & \cdots & -\alpha_n
\end{pmatrix}
$$
Then again $X=A+UCV.$ This time we get
$$
C^{-1}+VA^{-1}U = 
\frac{1}{n}\begin{pmatrix}
0 & -n & 0 \\
1 & n+1+\sum\limits_{i=2}^{n}\frac{1}{\alpha_i^2} &
\sum\limits_{i=2}^{n}\frac{1}{\alpha_i} \\
0 & -\sum\limits_{i=2}^{n}\frac{1}{\alpha_i} & 1
\end{pmatrix}
$$
As $\det(C^{-1}+VA^{-1}U)=\frac{1}{n^2}$, $X$ is always invertible if exactly one of the $\alpha_i$ is $0.$
A: The invertibility depends on the coefficients. For instance, when $n=3$ you have that the determinant is $$3(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)^2.$$
