# M is a square martix, and $φ^*$ : $V^* \to V^*$, then why $M^T=M_{φ^*_M}$? [duplicate]

Let M be a matrix, transpose $M^T$ is the matrix who's i,j component is the j,i component of M.

And let $\phi\in Hom(V,V)$, we define $\phi^*: V^* \to V^*$ by $\phi^*(f)(v)=f(\phi(v))$ where $V^*$ is the dual space of $V$.

$M_{φ^*_M}$ is the matrix representation of $φ^*_M$

Then how to prove $M^T=M_{φ^*_M}$

## marked as duplicate by amd, André 3000, JMP, Smylic, user91500Apr 30 '17 at 9:16

• You should probably define what $M_{\varphi_M^*}$ is. I assume it's the matrix representation of $\varphi^*$? – André 3000 Apr 30 '17 at 0:05
Fix a basis $e_1, \ldots, e_n$ of $V$, and let $e_1^*, \ldots, e_n^*$ denote the dual basis of $V^*$, defined by $$e_i^*(e_j) = \begin{cases} 1 & \text{if } i = j\\ 0 & \text{otherwise.} \end{cases}$$ Let $A$ be the matrix of $\varphi$ with respect to this basis, so the $i,j$ entry of $A$ is $a_{ij}$ where \begin{align} \label{A} \varphi(e_j) = \sum_i a_{ij} e_i \tag{1} \end{align} for all $j$. Similarly, let $B$ be the matrix of $\varphi^*$ with respect to the dual basis, so the $i,j$ entry of $B$ is $b_{ij}$ where \begin{align} \label{B} \varphi^*(e_j^*) = \sum_i b_{ij} e_i^* \, . \tag{2} \end{align} Applying the functional $\varphi^*(e_j^*)$ to the basis vector $e_k$, by (\ref{A}) we have \begin{align*} (\varphi^*(e_j^*))(e_k) &= (e_j^* \circ \varphi)(e_k) = e_j^*(\varphi(e_k)) = e_j^*\left(\sum_i a_{ik} e_i\right) = \sum_i a_{ik} e_j^*(e_i) = a_{jk} \, . \end{align*} On the other hand, by (\ref{B}) we also have \begin{align*} (\varphi^*(e_j^*))(e_k) = \left(\sum_i b_{ij} e_i^*\right)(e_k) = \sum_i b_{ij} e_i^*(e_k) = b_{kj} \, . \end{align*} Thus $a_{jk} = b_{kj}$ for all $j,k$, hence $B = A^T$.
• For clearness, should the question be ${M_{φ_M}}^T=M_{φ^*_M}$ ? – Parting Apr 30 '17 at 2:18
• Yes, or probably just ${M_\varphi}^T = M_{\varphi^*}$. I'm not sure what the second $M$ adds. – André 3000 Apr 30 '17 at 2:29