Let $V$ be a vector space over some field $k$ and $T \in \mathrm{GL}(V)$. Then, we can view $T\in \mathrm{GL}(\mathrm{Sym}^k(V))$ where $\mathrm{Sym}^k(V)$ denotes the $k^\mathrm{th}$ symmetric power of $V$ and denote it $T\_k$. Knowing $\det T$, is there a general formula for $\det T\_k$?



Determinant and symmetric power
We have that $\det T\_k$ is a fixed (depending on $n=\dim V$ and $k$ only) power of
$\det T$. To see this, as well as getting the power, one can for instance note
that $\mathrm{SL}(V)$ is the commutator subgroup of $\mathrm{GL}(V)$ (except for extremely small finite fields but we can always increase the size of the field) and hence
if $\det T=1$ then $\det T\_k=1$. We can then write any $T\in\mathrm{GL}(V)$ in
the form $DS$, where $S\in\mathrm{SL}(V)$ and $D$ a diagonal matrix with
diagonal entries $(t,1,1,\ldots,1)$. Then $\det((DS)\_k)=\det(D\_k)\cdot1$ so it
suffices to compute $\det(D\_k)$ but in the standard basis of $\mathrm{Sym}^kV$,
given a basis $e\_1,\ldots,e\_n$ of $V$, $D\_k$ is a diagonal entries and its
determinant is $t^R$, where $R=\sum\_{0\leq i\leq k}is^{n-1}\_{k-i}$. Here
$s^{a}\_{b}=\dim \mathrm{Sym}^bU$ where $\dim U=a$ which equals $\binom{a+b-1}{b}$.


