This is a very small question but I can't seem to shake it. In the lecture notes here it talks about discrete fourier transforms near the bottom. If we are letting $F_n$ be the $n$ by $n$ Fourier matrix. We know that $F_n$ is orthogonal and that
$$\frac{1}{\sqrt{n}}F_n\qquad (1)$$
is unitary. Because of this we have that the inverse of the above is $(n)^{-1/2}F_n^H=(n)^{-1/2}\overline{F}_n$ (we only need to take the conjugate because $F_n^T=F_n$), but the notes say that
$$\left(\frac{1}{\sqrt{n}}F_n\right)\left(\frac{1}{\sqrt{n}}F_n\right)=I$$
i.e. they drop the conjugation and say that $(1)$ is its own inverse. They gave an example of this and it worked for $F_4$ but I can't see why this would be generally true or if it even is. I'm sure if it is true then its specific to Fourier matrices and an explanation of what property of Fourier matrices makes it true would be nice.
It's exact quote is
Because $\frac{1}{\sqrt{n}}F_n$ is unitary, multiplying by $F_n$ and dividing by the scalar $n$ inverts the transform [the transform $F_n$].