# Textbook 2.5 (Qiskit) - Unitary and Hermitian matrices

In section 2.5 of the Qiskit textbook, it states that $$X$$, $$Y$$, $$Z$$ and $$H$$ are examples of unitary Hermitian matrices. As I understand it, this means that the following rule applies: $$UU^\dagger=U^\dagger U=1$$ I can't get the second part of this to work. Taking $$X$$ as an example: $$\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}.\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$ We can immediately see that $$U U^ \dagger=U^\dagger U$$ (as they're the same), but when I do an inner product, I get: $$(0\times0)+(1\times1)+(1\times1)+(0\times0)=2$$ Am I doing my matrix multiplication wrong, or am I forgetting something?

• Why are you doing an inner product? (and what are you doing it of?) You should be doing matrix multiplication and getting the answer of a $2\times 2$ matrix. Mar 2 at 12:24
• I was trying to get to a single scalar value, as that's what the equation says in the textbook. I think I was confusing my approach with the mechanism for calculating the probability of reading a qubit in a specific state. @ie-irodov 's answer below makes sense if I read the "1" as an "I". Mar 2 at 18:42
• It's one of the ways that people end up writing identity when writing for the web and using mathjax for latex rendering, which doesn't have the full set of Latex symbols we might otherwise be used to. It's just a shame it's not consistent with elsewhere in the book where they use I. Mar 3 at 8:00
• Yes - I think if the book had been habitually using "1" everywhere, I'd have been more comfortable! Thanks again. Mar 3 at 8:06

$$UU^{\dagger}=I$$ rather than $$1$$. Simply multiply $$U$$ and $$U^{\dagger}$$ as follows:
$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (0\times0) + (1\times1) & (0\times1)+(1\times0) \\ (1\times0) + (0\times1) & (1\times1) + (0\times0)\end{pmatrix}$$
$$= \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$$
It'll make more sense if you remember that appending $$U$$ and $$U^{\dagger}$$ one after the other in a circuit will make them equivalent to an identity gate, i.e. the net effect is cancelled out. For example, $$H|0\rangle=|+\rangle$$ and $$H|+\rangle=|0\rangle$$