Two level unitary matrices

In section 4.5.1 of Nielsen and Chuang, two-level unitary matrices are defined as unitary matrices which act non-trivially only on two or fewer vector components. I'm not sure that I understand this definition. For instance, is $$\begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\quad$$ a two level unitary matrix? It only acts non-trivially on the first and third vector components, but it doesn't leave this linear space invariant, unlike the examples given in Nielsen and Chuang.

• "It only acts non-trivially on the first and third vector components..." That's not correct. Enumerating the basis vectors as $\{e_1, e_2, e_3\}$, that matrix transforms $T e_2 = e_3$, which is not "trivial". – DanielSank Aug 4 '18 at 17:36
• Yes, I guess that acting trivially on component $j$ means that $Te_j = e_j$ I thought that it might mean that $U_{ij} = 0$ or $1$. I have continued reading and in the end you prove that unitaries such that $Te_j = e_j$ except for at most two j are universal, which is the aim of the section, so I guess that the definition you suggest is the right one. Thanks! – Alvaro Ortega Aug 4 '18 at 18:27

But, the following matrix works, i.e. is a two-level unitary matrix: $$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix}\quad$$
• Oops sorry. Well that can be fixed by changing any of the $\frac{1}{\sqrt{2}}$ into $-\frac{1}{\sqrt{2}}$ – Alvaro Ortega Aug 4 '18 at 18:15
Unitary matrices must satisfy $U^\dagger = U^{-1}$. Your example is not unitary because it isn't invertible so $U^{-1}$ doesnt exist. In that entire section, he is only talking about unitary (and therefore invertible) matrices.