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?
 A: 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$.
