I am reading about the Hilbert-Mumford criterion and I am stuck at something that is stated as "obvious" in every text that I can find. A bit of help would be much appreciated.
So, let $X$ be a projective variety over a field $k$, $G$ a reductive algebraic group acting on $X$ linearly, $\lambda : \mathbb{G}_m \rightarrow G$ a 1-parameter subgroup, and denote by $\lambda_x : \mathbb{G}_m \rightarrow X$ the map sending $t \mapsto \lambda(t)\cdot x$. Then this map extends to $\mathbb{P}^1$ (by the valuative criterion) - call the extended map $f$. Then Mumford denotes in his book the point $f(0)$ by $\lim_{t \rightarrow 0} \lambda(t)\cdot x$, and says that "clearly" it is fixed by the action of $\lambda(\mathbb{G}_m)$. I can understand this intuitively as follows: let $t,t' \in \mathbb{G}_m$ then $$\lambda(t) \cdot f(t') = \lambda(t)(\lambda(t') \cdot x) = \lambda(tt')\cdot x = f(tt')$$
Letting $t' \rightarrow 0$ we get that $f(0)$ is fixed, but I'd really like an algebraic proof for this. One can't just plug in $0$ into the above equations since $0 \not \in \mathbb{G}_m$ which is where I am stuck.
Also, they proceed to say that $f(0)$ being fixed by the action of $\lambda(\mathbb{G}_m)$ implies that $\lambda(\mathbb{G}_m)$ acts on the fiber of the line bundle (associated to the sheaf) $\mathcal{O}_X(1)$ at $f(0)$ by a character $t \rightarrow t^r$ for some $r \in \mathbb{Z}$. What I know is that since the action on $X$ is linear the projection map from the line bundle is equivariant so we must have that the action is an automorphism of the fiber, but how is this a charater?
Thanks.
