Is aI + bA invertible if rank(A) = n-1?

Question

I am not able to prove this for sure by myself... To be more precise, $A$ is a $n \times n$ matrix of rank $n-1$ such that all diagonal elements of A are positive, off-diagonal elements can be positive or negative, and its last row is the weighted sum of all the other rows ($i.e. A(n,:) = -1/n \sum_{i=1}^{n-1} i*A(i,:)$), which explains why $rk(A) = n-1$. I would then like to know if is possible to say that the matrix $\alpha I + \beta A $ with $\alpha >0$ and $\beta >0$ is invertible ? It seems to me that this is the case in general, though I could not proof that it was true for any $\alpha >0$ and $\beta >0$, any idea ?

Answer 1

The matrix $\alpha I+\beta A$ is invertible iff $-\alpha/\beta$ is not an eigenvalue of $A$.