Sesquilinear form, tensor product, vector space isomorphism

linear algebra

Let $V$ be a $n$-dimensional unitary vector space with $n < \infty$. Let $(e_1, \ldots, e_n)$ be an ON-basis of $V$ with dual basis $(e_1^*, \ldots, e_n^*)$. Define a sesquilinear form in $V \otimes V^*$ with

$\langle e_i \otimes e_j^*, e_k \otimes e_l^* \rangle _{V \otimes V^*} := \delta _{ik}\, \delta _{jl}$

Define a vector space isomorphism with

$\Phi _V \colon M(n \times n; \mathbb {C}) \to V \otimes V^*\,, \quad T \mapsto \sum _{i,j = 1}^n T_{ij}\, e_i \otimes e_j^*\,.$


Which of the following identities is true for all $S,T \in M(n \times n; \mathbb {C})$?

a. $\langle \Phi _V(S), \Phi _V(T) \rangle _{V \otimes V^*} = \mathrm {tr}(\overline {T} S)$

b. $\langle \Phi _V(S), \Phi _V(T) \rangle _{V \otimes V^*} = \mathrm {tr}(T^t S)$

c. $\langle \Phi _V(S), \Phi _V(T) \rangle _{V \otimes V^*} = \mathrm {tr}(\overline {T}^t S)$

Best Answer

Choice c is correct, assuming that $\bar T^t$ refers to the conjugate of the transpose of $T$. To see that this is the case, it suffices to note that the form is conjugate linear in the second argument and to plug in $T = v_i v_j^T$, where $v_1,\dots,v_n$ denotes the canonical basis of $\Bbb C^{n}$.

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