I have a unitary element $u\in C(\mathbb{T},M\_{n}(\mathbb{C}))$ such that $Spec(u)=\mathbb{T}$. Does there exist a unitary $v\in C(\mathbb{T},\mathbb{C})$ such that $Spec(uv)\subsetneqq\mathbb{T}$? 



Unitary with full spectrum
No. For example, let
$$ u(z) = \left(\begin{array}{cc} 1 & 0 \newline 0 & z \end{array}\right). $$
To see that for any $v \in C(\mathbb T, \mathbb C)$, the spectrum of $uv$ is $\mathbb T$, observe that
$$ uv(z) = \left(\begin{array}{cc} g(z) & 0 \newline 0 & h(z) \end{array}\right), $$
where the number of times that $g(z)$ winds around the circle (ie.\ its $K\_1$-class) is exactly one less than that of $h(z)$.
Thus, at least one of them winds around the circle a non-zero number of times, which implies that the spectrum of $uv$ is $\mathbb T$.


