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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?

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