# Unitary with full spectrum

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 $Spec(uv)\subsetneqq\mathbb{T}$?

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Care to add some more details about why you think this should be true, what you have tried, how it fits with all your previous Cstar questions, etc? –  Yemon Choi May 26 '13 at 23:02
It is not related to my previous questions. I don't know if this is true. But, if it holds then $v$ is homotopic to $u^{\ast}$. –  David May 26 '13 at 23:24

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 $\mathbb T$.