Suppose that we have linear transformation $T:V \to U$, then $T$ induces a map on dual spaces $T^*: U^* \to V^*$ via the pullback and suppose further that they are given by the matrices $A$ and $A^T$ respectively.
A matrix that is often of interest is $A^TA$. Since matrix multiplication corresponds to composition of linear maps, I would naively think that $A^TA$ corresponds to $T^* \circ T$ but this is not a valid composition (codomain and domain do not match)! So my question is what linear map $A^TA$ corresponds to?
Thank you!
In a way $A^TA$ corresponds to the composition $T^*\circ T$, because you can identify $U$ with its dual $U^*$ by choosing a basis $\{u_1,...,u_n\}$ of $U$ and mapping $u_i$ to its "dual vector" $u_i^*$ defined by $u_i^*(u_j) = \delta_{ij}$, this is a non-canonical isomorphism, let's call it $\phi$, of $U$ and its dual $U^*$. It is the isomorphism we get by making a column vector to a row vector by transposing it, if we have coordinates, which are just a basis. This is also the isomorphism you get by the bilinear map you get by just using the "standard scalar product" on your coordinate vectors. Then $A^TA$ corresponds to the map $T*\circ \phi \circ T$ (Just write it down in coordinates and it is quite clear).
You can't really see this in the product $A^TA$ because there you have implicitly chosen a basis by working with matrices. One also nicely sees the difference between row and column vectors in this example.