So suppose $U_1,U_2,U_1 \neq cU_2$ are unitary operators in finite dimesional complex vector space and $\nu_1,\nu_2$ are two complex numbers. What should I require from $\nu_1,\nu_2$ to make $\nu_1U_1 + \nu_2U_2$ also unitary?
My attempt:
Let $U = \nu_1U_1 + \nu_2U_2$.
$$
UU^{\dagger} = ( |\nu_1|^2+|\nu_2|^2)I + 2Re(\nu_1\bar{\nu_2}U_1U_2^\dagger)
$$
I want it to be equal to $I$, so:
$$
Re(\nu_1\bar{\nu_2}U_1U_2^\dagger) = \frac{1- |\nu_1|^2-|\nu_2|^2}{2}I
$$
And here I stucked. Help please.
Best Answer
Consider \begin{align} U_1 = \begin{pmatrix} e^{i\theta_1} & 0\\ 0 & e^{i\theta_2} \end{pmatrix} \ \ \text{ and } \ \ U_2 = \begin{pmatrix} e^{i\varphi_1} & 0\\ 0 & e^{i\varphi_2} \end{pmatrix} \end{align} which are clearly unitary. Next, observe if \begin{align} U=v_1U_1+v_2U_2 \end{align} then \begin{align} U^\dagger U = (|v_1|^2+|v_2|^2)I + 2\operatorname{Re}(\bar v_1v_2U_1^\dagger U_2). \end{align}
If you want $U$ to be unitary, then like you said \begin{align} 2\operatorname{Re}(\bar v_1v_2U_1^\dagger U_2) = (1-|v_1|^2-|v_2|^2)I. \end{align} However, we see that \begin{align} 2\operatorname{Re}(\bar v_1v_2U_1^\dagger U_2)&= 2\operatorname{Re} \left(\bar v_1 v_2 \begin{pmatrix} e^{i(\varphi_1-\theta_1)} & 0\\ 0 & e^{i(\varphi_2-\theta_2)} \end{pmatrix} \right)\\ &= 2|v_1||v_2|\operatorname{Re} \left( \begin{pmatrix} e^{i(\varphi_1-\theta_1+\chi)} & 0\\ 0 & e^{i(\varphi_2-\theta_2+\chi)} \end{pmatrix} \right)\\ &= 2|v_1||v_2| \begin{pmatrix} \cos(\varphi_1-\theta_1+\chi) & 0\\ 0 & \cos(\varphi_2-\theta_2+\chi) \end{pmatrix} \end{align} where $\chi$ is the difference in phase of $v_1$ and $v_2$. Clearly, this can't be a multiple of the identity in general.
So in general, I don't think a special complex linear combination of unitary matrices will be unitary, unless $U_1$ and $U_2$ satisfy additional properties.
Additional: In short, the choice of $v_1$ and $v_2$ only gives you one degree of freedom. However, $U_1$ and $U_2$ can have multiple degrees of freedom.
Correction: eyeballfrog has pointed out my mistake which can easily be amended. Indeed, one has to be careful with complex numbers. We have to augment the matrices to a $3\times 3$ matrices as follows \begin{align} U_1 = \begin{pmatrix} e^{i\theta_1} & 0 & 0\\ 0 & e^{i\theta_2} & 0\\ 0 & 0 & e^{i\theta_3} \end{pmatrix} \ \ \text{ and } \ \ U_2 = \begin{pmatrix} e^{i\varphi_1} & 0 & 0\\ 0 & e^{i\varphi_2} & 0\\ 0 & 0 & e^{i\varphi_3} \end{pmatrix}. \end{align} Then the same argument will lead to the desired conclusion that complex combination of unitary matrices need not be unitary. More precisely, we see that \begin{align} 2\operatorname{Re}(\bar v_1v_2U_1^\dagger U_2)&= 2|v_1||v_2|\operatorname{Re} \left( \begin{pmatrix} e^{i(\varphi_1-\theta_1+\chi)} & 0 & 0\\ 0 & e^{i(\varphi_2-\theta_2+\chi)} & 0\\ 0 & 0 & e^{i(\varphi_3-\theta_3+\chi)} \end{pmatrix} \right)\\ &=\ 2|v_1||v_2| \begin{pmatrix} \cos(\varphi_1-\theta_1+\chi) & 0 & 0\\ 0 & \cos(\varphi_2-\theta_2+\chi) & 0\\ 0 & 0 & \cos(\varphi_3-\theta_3+\chi). \end{pmatrix} \end{align}