Here is a theorem from representation theory:
$G$ is a reductive algebraic group in characteristic not equal $p$ and $E$ is an elementary abelian subgroup of $G$.
Suppose $G$ is complex and view the Lie algebra of G as a $\mathbb{C}E$-module with character $\chi_L$. Then dim $C_G(E)=(1/|E|)\sum_{x\in E}\chi_L(x)$.
I am being silly here.
I considered $E$ generated by \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} in $PGL_4(\mathbb{C})$. The centraliser apprently has dimension 9. But the Lie algebra of $PGL_n(\mathbb{C})$ is $\mathfrak{sl}(n,\mathbb{C})$ of traceless matrices. So by the formula, we would get dimension $0$...
How do I get the dimension right using the formula?
So we have $G=PGL_4(\mathbb C)$, accordingly the Lie algebra is $V= \mathfrak{sl}_4(\mathbb C)$, the traceless $4\times 4$-matrices. Note that the representation of $G$ on $V$ is the adjoint, i.e. an element $x\in G$ (and by restriction, $x\in E$), represented by a matrix $A \in GL_4(\mathbb C)$, acts on $X \in V$ via
$$\rho(x): X \mapsto A^{-1}XA.$$
Now $\chi_L(x)$ is defined as the trace of this $\rho(x) \in End_{\mathbb C}(V)$. So the fact that the underlying vector space of the representation consists of traceless matrices is kind of irrelevant, we are looking at traces of operators on that ($15$-dimensional) space $V$.
Now since your group $E$ is of order $2$, consisting of the identity $id$ and the non-trivial element $b$ represented by your matrix $B=\pmatrix{-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$, everyting is quickly computed: Of course $\chi_L(id)=\dim(V)=15$, whereas $\rho(b)$ is seen to operate with eigenvalue $-1$ on the six-dimensional space $$\pmatrix{0&*&*&*\\*&0&0&0\\*&0&0&0\\*&0&0&0}$$ but with eigenvalue $1$ on the nine-dimensional $$\pmatrix{*&0&0&0\\0&*&*&*\\0&*&*&*\\0&*&*&*} \subset \mathfrak{sl}_4(\mathbb C)$$.
So $\chi_L(b) = 9\cdot 1+6\cdot(-1)=3$, and the formula gives out the answer you want.