Linear combination of unitary operators is unitary

Question

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.

Answer 1

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}

Answer 2

This is not in general possible in $\mathbb C^n$ for $n\ge 3$, but is possible in $\mathbb C^2$.

Without loss of generality, assume $\nu_1$ is real and $\nu_2 = r e^{i\theta}$. Since $U_1$ and $U_2$ are unitary, $U_1^\dagger U_2$ is also unitary. Thus it has eigenvalues of the form $e^{i\lambda_k}$ for real $k$ and there is a unitary matrix $Q$ such that $U^\dagger_1 U_2 = Q^\dagger D Q$, where $D_{jk} = \delta_{jk}e^{i\lambda_k}$. Then we have $$ \nu_1 U_1 + \nu_2 U_2 = U_1(\nu_1 I + \nu_2 U_1^\dagger U_2) = U_1Q^\dagger (\nu_1I + \nu_2 D) Q = U_1Q^\dagger[(\nu_1 + \nu_2e^{i\lambda_k})\delta_{jk}]Q. $$ Since $U_1$ and $Q$ are unitary, this matrix will be unitary if and only if the diagonal matrix $(\nu_1 + \nu_2e^{i\lambda_k})\delta_{jk}$ is unitary, which is true if and only if $|\nu_1 + \nu_2 e^{i\lambda_k}|^2= \nu_1^2 + r^2 + 2\nu_1r\cos(\theta + \lambda_k) = 1$ for all $k$. In particular, this means for every pair $\lambda_j,\lambda_k$, we have $\cos(\theta+\lambda_j) = \cos(\theta+\lambda_k)$. This gives us a single parameter $\theta$ to satisfy every pair of eigenvalues. For dimensions 3 and higher, this is clearly not possible in general. For $\mathbb C^2$, we can choose $\theta= -(\lambda_k+\lambda_j)/2$. The norm condition becomes $\nu_1^2 + r^2 + 2\nu_1r\cos[(\lambda_j-\lambda_k)/2] = 1$, and we can choose $$ \nu_1 = \sqrt{1-r^2\sin^2\left(\frac{\lambda_k-\lambda_j}{2}\right)}-r\cos\left(\frac{\lambda_k-\lambda_j}{2}\right) $$ to satisfy the norm condition for any $r\in [0,1]$.