Consider the linear map $f: \mathbb{C}^3 \to \mathbb{C}^3$ defined by the matrix $$\begin{pmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 0 & 1 & 2 \end{pmatrix}.$$ Compute the trace of $\text{Sym}^2 \left(f \right)$ and that of $\text{Sym}^3 \left(f \right)$.
Let $V$ be a $F$-vector space. In my notes we define $\text{Sym}^k_{F}(V):= V^{\otimes n}/R$ where $R$ is the subspace generated by $v \otimes w - w \otimes v$. Now let $f$ be an $F$-linear map and let $\lambda_1,\ldots,\lambda_k \in F$ be eigenvalues of $f$ with counted multiplicities. Then $\text{Sym}^{k}(f)$ is the map between $\text{Sym}^{k}(V)$ and $\text{Sym}^{k}(V)$ with eigenvalues $\lambda_{i_1},\ldots,\lambda_{i_k}$ where $i \in S(k,d)=\{i:1 \leqslant i_1 \leqslant i_2 \leqslant \cdots \leqslant i_k \leqslant d \}$.
If $A$ is a matrix, we define the trace of $\text{Sym}^k(A)$ to be $\sum_{i \in S(k,n)} \lambda_{i_1} \cdots \lambda_{i_k}$. According to the fundamental theorem of symmetric polynomials, it follows that $\text{tr} \left(\text{Sym}^k(A) \right)$ is completely determined by the characteristic polynomial of $A$.
I computed the characteristic polynomial to be $f(T)=-T^3+4T^2-6T+9=-(T-3)(T^2-T+3)$. I found the eigenvalues of the characteristic polynomial: $\lambda_1=3$, $\lambda_2= \frac{1}{2}(1+i \sqrt{11}) $, $\lambda_3=\frac{1}{2}(1-i \sqrt{11})$.
Where do I go from here? I'm having a little trouble computing this using the above definition. Pointers in the right direction very much appreciated.