The trace of $A\otimes A$ as a self-map of the symmetric square $S^2V\leq V\otimes V$

Question

Let $V$ be a finite-dimensional complex vector space and $A\in \text{End}(V) $. Then we may define $A\otimes A\in \text{End}(V\otimes V)$ via $$ (A\otimes A)(v\otimes w)=(Av)\otimes(Aw)$$ and extending linearly. It is then easy to check that $\text{tr}(A\otimes A) = (\text{tr}A)^2 $.

Now let $S^2V\leq V\otimes V$ be the symmetric square of $V$, that is $$ S^2V = \langle v\otimes w + w\otimes v : v,w\in V\rangle.$$ Then $A\otimes A$ acts on $S^2V$ and my question was if there is a general formula for the trace of this restriction.

For example, if $(v_i:1\leq i \leq n)$ is an eigenbasis of $A$, say with $Av_i=\lambda_i v_i$, then $$ (v_i\otimes v_j + v_j\otimes v_i : 1\leq i \leq j\leq n)$$ is an eigenbasis of $A\otimes A$ on $S^2V$ and we can compute $$ \text{tr}_{S^2V}(A\otimes A) = \sum_{1\leq i\leq j \leq n}\lambda_i \lambda_j = \frac{1}{2}\left( \left(\sum_i \lambda_i\right)^2 + \sum_i \lambda_i^2\right)=\frac{1}{2}\left((\text{tr}A)^2 + \text{tr}(A^2)\right).$$

Does this formula hold for arbitrary $A$ by virtue of a general endomorphism being diagonal? If so is there a direct proof avoiding such a density argument?

Answer 1

Ok I think I figured it out. Let $(v_i:1\leq i \leq n)$ be a basis of $V$ and $(\varphi_i : 1\leq i\leq n)$ its dual basis.

Then $$(v_i\otimes v_i : 1\leq i \leq n)\ \bigcup\ \left(\frac{v_i \otimes v_j + v_j \otimes v_i}{\sqrt2}:1\leq i < j \leq n\right)$$ is a basis of $S^2V$ whose dual is $$(\varphi_i\otimes \varphi_i : 1\leq i \leq n)\ \bigcup\ \left(\frac{\varphi_i \otimes \varphi_j + \varphi_j \otimes \varphi_i}{\sqrt2}:1\leq i < j \leq n\right).$$

So we compute \begin{align*} \text{tr}_{S^2V}(A\otimes A)&=\sum_{i=1}^n \langle\varphi_i\otimes \varphi_i|A\otimes A|v_i \otimes v_i\rangle + \frac{1}{2}\sum_{i<j}\langle \varphi_i \otimes \varphi_j + \varphi_j \otimes \varphi_i|A\otimes A| v_i \otimes v_j + v_j \otimes v_i\rangle\\ &= \sum_{i=1}^n \langle\varphi_i|A|v_i\rangle^2 + \sum_{i<j}\bigg(\langle\varphi_i|A|v_i\rangle\langle\varphi_j|A|v_j\rangle+\langle\varphi_i|A|v_j\rangle\langle\varphi_j|A|v_i\rangle\bigg) \\ &=\frac{1}{2}\left( \left( \sum_{i=1}^n \langle\varphi_i|A|v_i\rangle \right)^2 + \sum_{1\leq i,j\leq n } \langle\varphi_i|A|v_j\rangle\langle\varphi_j|A|v_i \rangle \right)\\ &= \frac{1}{2}\left( (\text{tr}A^2 ) + \text{tr}(A^2) \right), \end{align*}

where in the last step we use $ \sum_j |v_j\rangle\langle \varphi_j| = 1$.