Matrix function that gives a scalar

Question

I have the following function:

$$f(z) = z\vec{b}^T[I-zA]^{-1}\vec{1},$$

where $z$ is a complex scalar with $Re(z)<0$ (for simplicity, WLOG, we can take $z$ to be real), $b$ is a vector, $1$ is a vector of ones. $I$ is the identity matrix, and $A$ is some arbitrary matrix so that $I-zA$ is non-singular.

I want to show that $Re(f)<0$ for some values of $z$. How can I solve the equation $f(z)=0$? Some help would be appreciated.

Answer 1

So, assuming that $z \in \mathbf{R}$ we have

1. $z = 0 \implies f(z) = 0$
2. $z \neq 0 \implies b^T(I - zA)^{-1}\cdot \mathbf{1} = 0$

Now, using the formula of inversion of the sum of matrices (see, e.g. here) we get $$ b^T\left(I + \frac{z}{1 - \text{tr}(A)z}A\right)1 = 0 \implies b^T\cdot \mathbf{1} + \frac{z}{1 - \text{tr}(A)z}b^TA\cdot \mathbf{1} = 0. $$ Expressing $z$ from the latter equation we get that $$ z = \frac{b^T1}{b^T \cdot \mathbf{1} \text{tr}(A) - b^TA \cdot \mathbf{1}}. $$