Compute the trace of $\text{Sym}^2 \left(f \right)$ and that of $\text{Sym}^3 \left(f \right)$.

Question

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.

Answer 1

Note that $A$ is diagonalizable. In particular, there is a basis $\{x,y,z\}$ for $\Bbb C^3$ such that \begin{align*} Ax &= \lambda_1 x & Ay &= \lambda_2 y & Az &= \lambda_3 z \end{align*} Now, as Daniel points out, $\DeclareMathOperator{Sym}{Sym}\Sym^2\Bbb C^3$ has basis $\{x^2,xy,xz,y^2,yz,z^2\}$. We can compute $\Sym^2 A$ on this basis \begin{align*} A x^2 &= \lambda_1^2x^2 & A xy &= \lambda_1\lambda_2 xy & Axz &= \lambda_1\lambda_3 xz & A y^2 &= \lambda_2^2 y^2 & Ayz &= \lambda_2\lambda_3 yz & A z^2 &=\lambda_3^2z^2 \end{align*} This proves that $\Sym^2A$ is similar to the diagonal matrix $$ \begin{bmatrix} \lambda_1^2 & 0 &0 &0 &0 &0\\ 0 & \lambda_1\lambda_2 & 0 &0&0&0\\ 0&0&\lambda_1\lambda_3 & 0 & 0 &0\\ 0&0&0&\lambda_2^2 & 0 & 0\\ 0&0&0&0&\lambda_2\lambda_3 & 0 \\ 0&0&0&0&0&\lambda_3^2 \end{bmatrix} $$ Thus $$ \DeclareMathOperator{tr}{tr}\tr\Sym^2 A = \lambda_1^2+\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2^2+ \lambda_2\lambda_3+\lambda_3^2 $$ Since \begin{align*} \lambda_1 &= -\frac{1}{2} i \, \sqrt{11} + \frac{1}{2} & \lambda_2 &= \frac{1}{2} i \, \sqrt{11} + \frac{1}{2} & \lambda_3 &= 3 \end{align*} we have $$ \tr\Sym^2 A =10 $$

Answer 2

Okay, so if $V = {\rm span}\{x, y, z\}$, then $${\rm Sym}^2 V = {\rm span}\{x^2, x y, x z, y^2, y z, z^2\}$$ Since $$A \cdot x = x + 2 y,$$ we have $$({\rm Sym}^2 A) \cdot x^2 = (Ax)(Ax) = (x + 2y)^2 = x^2 + 4 x y + 4 y^2$$ and so forth. Proceeding in this way, you can write a matrix for $({\rm Sym}^2 A)$, from which you can calculate the trace. And similarly for $({\rm Sym}^3 A)$.

Yes, there's a more sophisticated way of doing it, but until you understand what ${\rm Sym}^k V$ and ${\rm Sym}^k A$ are, you should do things the straightforward way first. Then once you understand what the things you're working with are you can go back and see why the theorem works.