Matrix of T'

Remark.

We can show mechanically that the matrix of $T$ is the transpose of the matrix of $T^{\prime }$. Suppose $e_1,...,e_n$ forms an orthonormal basis of $V$ and ${\phi }_1,...{\phi }_n$ forms the dual basis of $V^{\prime }$ (Duality). We use the definition of a matrix: $M(T)=(a_{ij})$ if $T{e}_j={\sum }_{i=1}^n{a}_{ij}{e}_i$. Observe that for any basis vector $e_j$,

$$T^{\prime }({\phi }_i(e_j))={\phi }_i\circ T(e_j)={\phi }_i(\sum _{k=1}^n{a}_{kj}{e}_k)=a_{ij}.$$

The following linear combination of dual basis vectors acts the same way on the basis vectors of $V$:

$$(\sum _{k=1}^n{a}_{ik}{\phi }_k)(e_j)=a_{ij}.$$

Since linear maps are uniquely determined by how they act on a set of basis vectors, it must be true that $T^{\prime }{\phi }_i={\sum }_{k=1}^n{a}_{ik}{\phi }_k$. Equivalently, $T^{\prime }{\phi }_j={\sum }_{i=1}^n{a}_{ji}{\phi }_i$ where we simply replace $i$ with $j$ and $k$ with $i$. This implies that $M(T^{\prime })=(a_{ji})$.

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This result took me a while to wrap my head around and gain an intuition. For me, what made it click was the realization that $T^{\prime }{\phi }_i={\sum }_{j=1}^n{a}_{ji}{\phi }_j$ is simply a unique linear combination of the dual basis vectors.