What is the explicit matrix for tensors $V \otimes V^*$

Question

Notice the natural isomorphism $V \otimes V^* \simeq End(V)$, I wonder what is the explicit matrix given by $v \otimes v^*$? (where $v^* \in V^*$ is the dual space of V).

Answer 1

Fix a basis $\mathcal{B}^* = (e_1,\ldots, e_n)$ for $V$, and write $v = \sum_{i=1}^n a^ie_i$. If $v^* = \sum_{j=1}^n b_je^j$, where $(e^1,\ldots, e^n)$ is the dual basis to the original one chosen, recalling that $(v\otimes v^*)(w) = v^*(w)v$, we compute the matrix of $v\otimes v^*$ with the usual algorithm: evaluate it at $e_j$, write it as a combination of the original basis, and place the coefficients in columns. We have $$(v\otimes v^*)(e_j) = v^*(e_j)v = b_jv = \sum_{i=1}^n a^ib_je_i.$$So the matrix is $[v\otimes v^*]_{\mathcal{B}} = (a^ib_j)_{i,j=1}^n$. If $[v]_{\mathcal{B}}$ is the column vector consisting of the components of $v$ relative to $\mathcal{B}$ and $[v^*]_{\mathcal{B}}$ is the column vector consisting of the components of $v^*$ relative to $\mathcal{B}^*$, then $$[v\otimes v^*]_{\mathcal{B}} = [v]_{\mathcal{B}}[v^*]_{\mathcal{B}}^\top.$$