At "Complex Hadamard Matrices", I found that, two Kronecker (tensor) products of Fourier matrices $k_1$ and $k_2$ are equivalent, if and only if $k_2$ can be obtained from $k_1$ by a combination of an arbitrary number of operations from the set:
- changing the order of Kronecker product factors,
- replacing a subproduct $F_{m(k)}⊗F_{m(k+1)}$ by the factor $F_{m(k)⋅m(k+1)}$ only if $m(k)$ and $m(k+1)$ are relatively prime,
- replacing a factor $F_{ab}$ by the subproduct $F_a⊗F_b$ only if $a$ and $b$ are relatively prime [Ta05].
Especially the last two bullets reminded me to what I've read here "Multiplicative function":
In number theory, a multiplicative function is an arithmetic function $f(n)$ of the positive integer $n$ with the property that $f(1) = 1$ and whenever $a$ and $b$ are coprime, then $$ f(ab) = f(a)\cdot f(b). $$
My Question:
Since $F_1=1$, does this mean that the Fourier matrix $F_n$ represents kind of a (tensor) multiplicative function?