I have $ n \times n $ real matrix $A$ with 1 on the diagonal and a constant $ c \ne 1 $ on non diagonal entries. I want to prove the matrix is invertible or come with a counter example. I think it is true and tried many examples but I'm not sure how to prove it.
Edit: I require that $ c \ne -1 $ as well
Edit : $ 0<c<1 $
In this case I think $A$ is invertible : define $c=cos^2\alpha $ and have $ A=sin^2\alpha I_n+cos^2\alpha J_n $ where $J_n$ is the all one matrix. $\; sin^2\alpha I_n $ is positive definite and $cos^2\alpha J_n $ is PSD (leading principle minors are non negative). Their sum must be positive definite so $A$ is positive definite and therefore invertible.
Is this a valid proposition ?
Thanks!
Best Answer
The all-1's matrix $J$ of rank $1$ has eigenvalue $0$ with multiplicity $n-1$ by the rank nullity theorem, and $n$ with multiplicity $1$ (the all-1's vector is an eigenvector).
Your matrix is $cJ+(1-c)I$. Its eigenvalues, then, are $(1-c)$ with multiplicity $n-1$ and $cn+(1-c)$ with multiplicity $1$.
Your matrix is non-singular provided these eigenvalues are non-zero.