If I have two unitary matrices $U_1, U_2$, would the following be true for $\lambda \in [0, 1]$?
$$
((1 – \lambda) U_1 + \lambda U_2)((1 – \lambda) U_1 + \lambda U_2)^\dagger = ((1 – \lambda) U_1 + \lambda U_2)^\dagger((1 – \lambda) U_1 + \lambda U_2) = \mathbb{I},
$$
where $\mathbb{I}$ is the identity matrix of the appropriate dimension. Here is my attempt to prove it:
$$
((1 – \lambda) U_1 + \lambda U_2)((1 – \lambda) U_1 + \lambda U_2)^\dagger \\
= ((1 – \lambda) U_1 + \lambda U_2)((1 – \lambda) U_1^\dagger + \lambda U_2^\dagger) \\
= (1 – \lambda)^2 U_1 U_1^\dagger + \lambda(1 – \lambda) U_1U_2^\dagger + \lambda(1 – \lambda) U_2U_1^\dagger + \lambda^2 U_2 U_2^\dagger \\
= (1 – \lambda)^2 \mathbb{I} + \lambda(1 – \lambda) (U_1U_2^\dagger + U_2U_1^\dagger) + \lambda^2 \mathbb{I} \\
= ((1 – \lambda)^2 + \lambda^2) \mathbb{I} + \lambda(1 – \lambda) (U_1U_2^\dagger + U_2U_1^\dagger)
$$
I got stuck here. Any help would be highly appreciated. TIA.
Best Answer
False. Take $A=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$, which is orthogonal, then $A+A^{T}=0$, that has no inverse, and then your postulate fails for example for $\lambda=1/2$ because $A^{T}$it is also an orthogonal matrix.