Is this monotonic in $a,b$? I want to claim that $$\det(I+[aA,bB]VV^*[aA,bB]^*)$$ is increasing in both $a$ and $b$. where $V$ is a complex valued matrix so are $A$ and $B$. $a,b$ are real positive and $^*$ denotes conjugate transpose (Hermitian). $[aAbB]$ represents the augmentation of the the matrices by putting them side by side in that order.
Also in case the claim is wrong, for what $V$ is it valid? I can see that it holds for real valued vector for example $[aA,bB]=(a,b)$ and $V=(c,d)$ where all values are reals. Also holds for $a$ or $b$ zero in matrix case. I think it may have a solution in eigenvalues of positive semi-definite matrices (Is $\det(I+aAVV^*A^*)$ increasing function in $a$.). It holds for $a=b$ as well.
PS. NOT home work, my research in information theory.
Thank you.
 A: Let us denote
$$V = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}.$$
Then
$$\begin{bmatrix} aA & bB \end{bmatrix} V = aAV_1 + bBV_2,$$
so
$$I + \begin{bmatrix} aA & bB \end{bmatrix} V V^* \begin{bmatrix} aA & bB \end{bmatrix}^* = I + (aAV_1 + bBV_2)(aAV_1 + bBV_2)^*.$$
Using the above, I made this in Mathematica:
n=2;
V1=IdentityMatrix[n];
V2=IdentityMatrix[n];
A={{0,1},{1,0}};
B=-2IdentityMatrix[n];
aAV1bBV2[a_,b_]:=a A.V1+b B.V2;
d[a_,b_]:=Det[
    IdentityMatrix[n]+aAV1bBV2[a,b].ConjugateTranspose[aAV1bBV2[a,b]]
];
Plot3D[d[a,b],{a,0,2},{b,0,2}]
Plot[d[a,1],{a,0,2}]

Here is the manually rotated 3D graph (note the back side):

Here is the 2D graph for $b = 1$:

So, what happened? For
$$V = I, \quad A = \begin{bmatrix} & 1 \\ 1 \end{bmatrix}, \quad B = -2I,$$
using $A^2 = I$, we have:
\begin{align*}
X &:= I + \begin{bmatrix} aA & bB \end{bmatrix} V V^* \begin{bmatrix} aA & bB \end{bmatrix}^* = I + (aAV_1 + bBV_2)(aAV_1 + bBV_2)^* \\
&= I + (aA - 2bI)(aA - 2bI)^* = I + (aA - 2bI)^2 = I + a^2I - 4abA + 4b^2I \\
&= (1 + a^2 + 4b^2)I - 4abA = \begin{bmatrix}
1 + a^2 + 4b^2 & -4ab \\
-4ab & 1 + a^2 + 4b^2
\end{bmatrix}.
\end{align*}
For constant $b = 1$, determinant of that is
$$\det X = (1 + a^2 + 4b^2)^2 - (-4ab)^2 = (1 + a^2 + 4)^2 - 16a^2 = a^4 - 6 a^2 + 25.$$
Since
$$\frac{\Delta \det X}{\Delta a} = 4 a (a^2 - 3),$$
it is easy to see that this function has minimums at $\pm\sqrt{3}$.
Note: the same would happen for $B = 2I$, but I didn't expect such a simple example to counter your statement.
As for the part "for what $V$ is the statement valid", I'd say for none. Assuming that $V_1$ and $V_2$ are nonsingular, you can easily construct a counterexample from the above example for any such $V$. There remains a case when either $V_1$ or $V_2$ are singular, but I think this can be overcome as well.
A: Suppose $A$ is $m\times n$ and $B$ is $m\times p$. Let $M=\begin{bmatrix}M_{11}&M_{12}\\ M_{12}^\ast&M_{22}\end{bmatrix}=\begin{bmatrix}A&0\\ 0&B\end{bmatrix}VV^\ast\begin{bmatrix}A&0\\ 0&B\end{bmatrix}^\ast$, where all the four subblocks of $M$ are $m\times m$. Then
\begin{align*}
f(a,b):=&I+[aA|bB]VV^*[aA|bB]^*\\
=&I+[aI_m|bI_m]M\begin{bmatrix}aI_m\\ bI_m\end{bmatrix}\\
=&I+a^2M_{11}+2abH_{12}+b^2M_{22},
\end{align*}
where $H_{12}=\frac12(M_{12}+M_{12}^\ast)$ is the Hermitian part of $M_{12}$.
So, what matters is the matrix $M$ but not individual choices of $A,B$ and $V$. Using the derivative formula for determinant, we get
\begin{align*}
\frac{d}{da}\det f(a,b)
=\det f(a,b)\ \operatorname{trace}\left[f(a,b)^{-1}(2aM_{11}+2bH_{12})\right],\tag{1}\\
\frac{d}{db}\det f(a,b)
=\det f(a,b)\ \operatorname{trace}\left[f(a,b)^{-1}(2aH_{12}+2bM_{22})\right].\tag{2}
\end{align*}
Hence a necessary and sufficient condition for $\det f(a,b)$ to be monotonic in both $a$ and $b$ is that the two traces on the RHS of $(1)$ and $(2)$ are nonnegative whenever $a,b>0$. Since $f(a,b)$ is positive definite and $M_{11},M_{22}$ are always positive semidefinite, one obvious sufficient condition for monotonicity of $\det f$ is that $H_{12}$ is positive semidefinite, and one obvious necessary condition is that $H_{12}$ is not negative definite.
When $H_{12}$ is not positive semidefinite, $\det f(a,b)$ may fail to be monotonic in either variable. For a counterexample, consider $M=\begin{bmatrix}1&-2\\ -2&4\end{bmatrix}$ (which can be constructed using $A=B=1$ and $V^\ast=(1,-2)$). In this case, $\det f(a,b)=1+(a-2b)^2$, which is neither monotonic in $a$ nor monotonic in $b$ at any point on the line $a=2b$.
