Defining Segre classes of a vector bundle using projective bundle of hyperplanes

Question

Let $X$ be an algebraic scheme over a field and $E$ a vector bundle of rank $e+1$ on $X$. Denote by $\mathbf{P}(E) = \mathrm{Proj}(\mathrm{Sym}(E))$ the projective bundle of hyperplanes in $E$ and by $P(E) := \mathbf{P}(E^\vee)$ the projective bundle of lines in $E$. In Fulton's intersection theory, the latter is used to define intersection with the Segre classes $s_i(E)$. Namely if $p : P(E) \to X$ denotes the projection and $p^*E^\vee \to O_{P(E)}(1)$ is the universal quotient line bundle, then he defines $$s_i(E) \cap \alpha := p_*(c_1(O_{P(E)}(1))^{e+i} \cap p^*\alpha)$$ for $\alpha \in A_*(X)$. I wonder if one can also use the bundle $\mathbf{P}(E)$ to define Segre classes. In other words, denote by $p' : \mathbf{P}(E) \to X$ the projection and by $p'^* E \to O_{\mathbf{P}(E)}(1)$ the unversal quotient line bundle. Define $$s_i'(E) \cap \alpha := p'_*(c_1(O_{\mathbf{P}(E)}(1))^{e+i} \cap p'^*\alpha)$$ for $\alpha \in A_*(X)$. Will we have $s_i(E) = s_i'(E)$?

Answer 1

The quantity $s_i'(E)$ is by definition $s_i(E^\vee)$. A standard fact is that $s_i(E^\vee) = (-1)^i s_i(E)$. This is seen for example by comparing the Chern roots.