Prove that matrix is invertible

Question

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

Answer 1

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 $0.$

Answer 2

The invertibility depends on the coefficients. For instance, when $n=3$ you have that the determinant is $$3(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)^2.$$