If $\xi$ is a rank 1 element in $K(X)$ for some compact topological space $X$, that is a virtual bundle whose virtual dimension is 1 at any point of $X$, then is it invertible in the ring $K(X)$?
Here $K$ stays for complex topological $K$-theory and $X$ is a compact Hausdorff space.
Not necessarily. We have
$$ K(S^0) \cong \mathbb{Z} \oplus \mathbb{Z} $$
where the product is component-wise. Then the element $(3,-2)$ has virtual dimension one, but it is not invertible.
However, the answer is yes if the space $X$ is connected. Consider the augmentation map $\epsilon \colon K(X) \to \mathbb{Z}$ which sends a virtual bundle to its virtual dimension and let us denote by $1$ the trivial $1$-dimensional bundle over $X$. A virtual bundle $c$ has virtual dimension $1$ if and only if $c-1$ belongs to the kernel of $\epsilon$. This kernel is $\widetilde{K}(X)$ and any element here is nilpotent, so $c-1$ is nilpotent. Therefore $c = 1 + (c-1)$ is invertible.
Note: Any element in $\widetilde{K}(X)$ is nilpotent by Corollary 3.1.6 in Atiyah's K-theory book. The group $K_1(X)$ from this corollary coincides with $\widetilde{K}(X)$ when $X$ is connected, see pages 120-121 in the same book.