Let’s pick up our discussion of matrices and forms and try to tie both views of our matrix together. We’re considering a bilinear form on a vector space over the real or complex numbers, which we can also think of as a linear map from to its dual space . We can also consider a sesquilinear form on a complex vector space, which is equivalent to an antilinear (conjugates scalars) map from the space to its dual.
We got a matrix by picking a basis for and plugging basis vectors into our form
And we also found that this is the matrix of the map from to , written in terms of the basis and its dual basis .
Now we have two bases of the same cardinality, so there is a (highly non-canonical) linear isomorphism from to which sends the basis vector to the basis vector . If we compose this with the map from to given by the form , we get a linear map from to itself, which we will also call . If is sesquilinear, we use the unique antilinear isomorphism sending to , and again get a linear map . The matrix of this linear map with respect to the basis is , just as before.
This seems sort of artificial, but there’s method here. Remember that we can come up with an inner product by simply declaring the basis to be orthonormal. Then the linear functional is given by . It’s this correspondence that is captured by both the non-canonical choice of inner product, and by the non-canonical choice of isomorphism.
But now we can extract the same matrix in a slightly different way. We start by hitting the basis vector with the new linear transformation to come up with a linear combination of the basis vectors. We then extract the appropriate component by using the inner product. In the end, we find
So in the presence of a given basis (or at least a given inner product) any bilinear (or sesquilinear) form corresponds to a linear transformation, and the matrix of the linear transformation with respect to the selected basis is exactly that of the form itself. On the other hand, if we have a linear transformation from to itself, we can stick it into the inner product as above and get a bilinear (or sesquilinear) form out.