For definition of collineation, please check this link
An axis of a collineation $f$ is a line of a projective plane such that $f$ fixes all points on that line and say $f$ is axial iff $f$ has an axis. $f$ is central iff there is a point $P$ on the projective plane such that any line passing through $P$ will be fixed by $f$ (hence $P$ is fixed as well) and in this case we call $P$ a centre. Now we want to show that a collineation will have an axis if having a centre.
Want $\exists$ a line $l$ such that $\forall p \in l, f(p) = p$.
Assume by contradiction: $\forall$ $l, \exists p \in l$ such that $f(p) \neq p$.
Fix a line $l$, define $D_l$ to be {$a \in l|a \neq f(a)$} and under our assumption, $D_l$ is non-empty for all $l$. For any two points on the plane, say $b, c$, call the line passing through them $l_{b, c}$. For example, say a line $l'$ has points $a, b, c, d$ and hence $l' = l_{b, c} = l_{a, d} = l_{a, c} = l_{d, a}$. Let $P$ be the given centre of the collineation $f$.
Claim: for any fixed line $l$ that does not pass through $P$, $\exists a \in D_l$ such that $P \notin l_{a, f(a)}$.
Verfication: Assume by contradiction, $P \in l_{a, f(a)} \forall a \in D_l$.
Fix $a \in D_l$ and since $l$ does not pass $P$ but $l_{a, f(a)}$ does, $f(a) \notin l$ and hence $f(l) \neq l$. As we mentioned in the definition, $f$ fix $P$ and hence $l_{a, f(a)}$ will not be $f(l)$ because $f$ preserve incidence. Say $c \in {f(l) \cap l}$ and then we notice that $l_{f(a), c} = f(l)$. Assume $c = f(d)$ for some $d \in l$.
a): if $d \neq f(d)$, then $l_{c, d} = l_{d, f(d)} = l$ and we have $d \in D_l$ and $l$ does not pass through $P$. Here is the contradiction.
b): Now I do not know how to deal with the case when $d = f(d)$ and feel like I might picked the wrong path since the beginning ....
Eventually, after I proved the claim, I want to fix a line $l$ that does not pass through $P$ and has $a \in D_l$ such that $P \notin l_{a, f(a)}$. Then $l_{a, P} \neq l_{f(a), P} = f(l_{a, P})$. Here I obtain the contradiction.
If my idea is correct, any hints on how to continue will be appreciated. Also, hints about another solution is welcomed and I am sure that there is an easier way on some other references (but I just somehow came across the current one .....)
This is an application of Desargues' Theorem.
Pick three points $A, B, C$ on three invariant lines which pass through the center and define $\bigtriangleup ABC$. The image $\bigtriangleup A'B'C'$ of $\bigtriangleup ABC$ will be centrally perspective with $\bigtriangleup A'B'C'$. Using Desargues' theorem, this implies that the intersections of $AB$ and $A'B'$, $BC$ and $B'C'$, and $AC$ and $A'C'$ will be collinear. These intersections are clearly fixed points under the collineation. So, as we have found three fixed points on a line, the whole line must be point-by-point invariant and hence must be the axis of the collineation.