I'm trying to determine if the matrix:
\begin{bmatrix}
0 & i \\
1 & 0
\end{bmatrix}
is a unitary matrix.
Therefore, the first step I'm taking is to figure out if both $\langle 0, 1\rangle$ and $\langle i, 0\rangle$ are unit vectors. $\langle 0, 1\rangle$ is obviously a unit vector, however, I'm not sure exactly what the rule is when it comes to determining if a vector is a unit vector when you have imaginary numbers in it, because at least in this case, the magnitude is not $1$, it is:
$$
\mathrm{mag}(\langle i, 0\rangle) = i
$$
(which is the sqrt of $-1$) and not $1$. So even though the magnitude is imaginary. Is it still considered a unit vector?
Best Answer
Yes, it is a unitary matrix, because $A^HA$ is the identity, where $A^H$ denotes the hermitian transpose: $$ A=\begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix} \qquad A^HA= \begin{bmatrix} 0 & 1 \\ -i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$