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

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?

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