I have an ODEs problem which confuses me.
The problem statement is:
Let $X(t)$ be a fundamental matrix of x'(t) = A(t)x(t),\qquad\qquad\qquad\qquad (1) where $A(t + \omega) = A(t)$ for some $\omega > 0$. Define $P(\omega)$ to be the set of all solutions to (1). Suppose $f\in C(\mathbb{R}\to\mathbb{R}^{n})$ is a function such that $f(t + \omega) = f(t)$ for all $t$. Show that the following are equivalent:
(1) The equation x'(t) = A(t)x(t) + f(t) has a unique solution.
(2) $X^{-1}(\omega) - X^{-1}(0)$ is non-singular.
(3) $\dim(P(\omega)) = 0$.
I think that $(1)\Leftrightarrow (3)$ is trivial. But $(2)$ confuses me. If $X$ is a fundamental matrix, its columns must each be solutions to x'(t) = A(t)x(t), which means the unique solution from the first condition wouldn't be unique since it could be added to any multiple of any column of $X(t)$ which would generate a new solution to x'(t) = A(t)x(t) + f(t).
The only way this problem would be avoided is if $X(t)$ had non non-zero columns which would make (2) a nonsense statement.
What's my logic error here?