Let $V$ be the space of all $3 \times 3$ antisymmetric matrices. And let $T \in T_{2}^{1}$ (tensors) be associated with the map given as $Z(A,B) = AB - BA$, where $A,B \in V$.
The basis of $V$ is given as a classic basis of antisymmetric matrices: $\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\!, \\[2pt]$$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\!, \\[2pt]$$\begin{pmatrix} 0 & 0& 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}\!, \\[2pt]$
Find $[T]_B$.
I just have no idea how to work with tensors given like that. Should I write matrix $A$ as an arbitrary antisymmetric matrix and then apply the $Z$ map? But how is this a tensor problem?
Thanks for any suggestions.
If your basis of antisymmetric matrices is given by $A_1, A_2, A_3$ and its dual basis on $V^*$ is denoted by $A_1^*, A_2^*, A_3^*$, then a basis of $T_2^1(V)$ is given by $$(A_k \otimes A_i^* \otimes A_j^*)_{k,i,j}.$$ I think that you are asked here to express $T$ in this basis.
Do you know how to do it from here?
The identification $T_2^1 (V) \cong \text{Bilinear maps}~\colon V × V → V$ is actually an isomorphism of linear spaces given by \begin{align*} T_2^1(V) &→ (\text{Bilinear maps}\colon V × V → V),\\ w \otimes φ \otimes ψ &↦ [(v,v') ↦ φ(v)·ψ(v')·w] \end{align*} Hence, if you want to express $T ∈ T_2^1(V)$ in terms of the basis above, you can instead express its corresponding bilinear map $Z$ in terms of the images of this bases, namely the bilinear maps $$α_{k,i,j} \colon V × V → V,~(v,v') ↦ A_i^*(v)·A_j^*(v')·A_k.$$ So now, there is some linear combination $$Z = \sum_{κ,ρ,σ} λ_{κ,ρ,σ}α_{κ,ρ,σ}$$ and you are interested in those $λ_{κ,ρ,σ}$. To get them, just plug in $A_i$ and $A_j$ – all summands except for $λ_{k,i,j}α_{k,i,j}$ for $k=1,2,3$ will vanish. So just express the result of $Z(A_i,A_j)$ in terms of $A_k$.
Summary: This all amounts to calculating $Z(A_i,A_j)$.