Are eigenvalues of $ (I-BA-B\phi)^{-1}(B-B\phi)$ smaller than those of $(I-BA)^{-1}B$? Notation and assumptions: denote by $\phi$ a scalar and its absolute value $|\phi|<1$. $B$ and $A$ are $n\times n$ matrices. All eigenvalues of $A$, $BA$, and $BA+B\phi$ have real parts less than one. $I$ stands for identity matrix.
Question: if all eigenvalues of $(I-BA)^{-1}B$ have real parts less than one, will we have that all eigenvalues of $(I-BA-B\phi)^{-1}(B-B\phi)$ have real parts less than one? 
I would appreciate very much if anyone could help. I think this may be true but cannot prove it. For scalar case, we have, if $\frac{a}{b}<1$, $c<a$ and $c<b$, then $\frac{a-c}{b-c}<1$. Do we have a counterpart for matrices?
 A: What I am providing is not a complete proof. But, it proves the assumption for a special case. So, it kinda backs up the idea.
Let's consider a special case where both $B$ and $A$ share the same set of eigenvectors. $\alpha_i$ for $i=1,2,...,n$ gives the eigenvalues of $B$ and $\beta_i$ gives the eigenvalues of $A$.
First, the eigenvalues of $(I-BA)^{-1}B$ are found, step by step.
1-Eigenvalues of $BA$ are $\alpha_i\beta_i$
2-Eigenvalues of $I-BA$ are $1-\alpha_i\beta_i$
3-Eigenvalues of $(I-BA)^{-1}$ are $\frac{1}{(1-\alpha_i\beta_i)}$
4-Eigenvalues of $(I-BA)^{-1}B$ are $\frac{\alpha_i}{(1-\alpha_i\beta_i)}$
If $\alpha_i$ and $\beta_i$ are real numbers, between zero and one, then $\frac{\alpha_i}{(1-\alpha_i\beta_i)}$ would be less than one if $\alpha_i<1-\alpha_i\beta _i$
Now, rewrite the second matrix as below
$(I-BA-\phi B)^{-1}(B-B\phi)=(I-B(A+\phi I))^{-1}(1-\phi)B=(1-\phi)(I-B(A+\phi I))^{-1}B$
Going the same procedure (steps) as for the previous one, the eigenvalues of this matrix are
$\frac{\alpha_i(1-\phi)}{1-\alpha_i(\beta_i+\phi)}=\frac{\alpha_i(1-\phi)}{1-\alpha_i\beta_i-\alpha_i\phi}$
To draw contradiction, let's assume
$\frac{\alpha_i(1-\phi)}{1-\alpha_i\beta_i-\alpha_i\phi}>1$
Then, it is easily seen that
$\alpha_i>1-\alpha_i\beta _i$
Which contradicts the inequality ($\alpha_i<1-\alpha_i\beta _i$) from the first matrix.
Special thanks to @Felix, for correcting a fatal error.
