Maybe, this is a problem of linear algebra. But I do not know how to calculate it. Let $E$ be a vector bundle of rank $2$ over an algebraic surface. $H=S^{2n}E\bigotimes (\operatorname{det} E)^{\bigotimes -n}$, then $\operatorname{det} H$ is trivial, why?

Maybe, this is a problem of linear algebra. But I do not know how to calculate it. Let $E$ be a vector bundle of rank $2$ over an algebraic surface. If $H=S^{2n}E\bigotimes (\operatorname{det} E)^{\bigotimes -n}$, then $\operatorname{det} H$ is trivial, why?

Maybe, this is a problem of linear algebra. But I do not know how to calculate it. Let $E$ be a vector bundle of rank 2 over an algebraic surface. $H=S^{2n}E\bigotimes (det E)^{\bigotimes -n}$, then $det H$ is trivial, why?

Maybe, this is a problem of linear algebra. But I do not know how to calculate it. Let $E$ be a vector bundle of rank $2$ over an algebraic surface. $H=S^{2n}E\bigotimes (\operatorname{det} E)^{\bigotimes -n}$, then $\operatorname{det} H$ is trivial, why?