Suppose $e_1,\dots,e_n$ is a basis of a vector space $V$ and $\epsilon_1,\dots,\epsilon_n$ be the dual basis of $V^*$. Let $\alpha\colon V^*\otimes V\to \text{Hom}(V,V)$ be the canonical isomorphism.
I am interested in finding an $x\in V^*\otimes V$ such that $\alpha(x)=\alpha(t)\circ\alpha(t)$, where $t=(2\epsilon_1-\epsilon_3)\otimes (e_1+e_2)$, but I am not sure how to go about this problem.
Any help/hint will be greatly appreciated. Thanks.
On
You can use the composition product $\circ$ in $V^*\otimes V$, which by definition makes $\alpha$ an algebra isomorphism and satisfies $$(u^*\otimes u)\circ(v^*\otimes v)=\langle u^*,v\rangle(v^*\otimes u)$$ where $\langle u^*,v\rangle=u^*(v)$.
It follows that $$(u^*\otimes u)\circ(u^*\otimes u)=\langle u^*,u\rangle(u^*\otimes u)$$
Taking $u^*=2\epsilon_1-\epsilon_3$ and $u=e_1+e_2$ so that $t=u^*\otimes u$, we have $\langle u^*,u\rangle=2$, so $x=t\circ t=2t$.
The canonical isomorphism satisfies $\alpha(f\otimes v)(w)=f(w)v$. So it's nothing more than a computation. If $v=\lambda_1e_1+\lambda_2e_2+\lambda_3e_3$ then:
$\alpha(t)(v)=(2\epsilon_1(v)-\epsilon_3(v))(e_1+e_2)=(2\lambda_1-\lambda_3)(e_1+e_2)=(2\lambda_1-\lambda_3)e_1+(2\lambda_1-\lambda_3)e_2$
From here it's easy to compute $\alpha(t)\circ\alpha(t)$, and that will immediately give the element $x$.