Rank of the harmonics in a Fourier series expansion

Question

Let $\boldsymbol{A}(t)$ be a $T$-periodic matrix with rank $r$, and $\boldsymbol{A}_n$ the harmonics of its Fourier series expansion, so that $$ \boldsymbol{A}(t) = \sum_{n=-\infty}^{+\infty} \boldsymbol{A}_n e^{\mathrm{i}n\frac{2\pi}{T} t}. $$ Also the matrices $\boldsymbol{A}_n$ will be of rank $r$?

Answer 1

This already fails in the $1$-dimensional case: a nonzero function may have some Fourier coefficients equal to zero.

Let $A(t) = I \,\exp(2\pi in t/T)$, where $I$ is the identity matrix of size $r$. Then the rank of $A(t)$ is $r$ for all $t$, but the coefficient $A_2$ is the zero matrix. (As are all coefficients except $A_1=I$).