## Matrices IV

Like we saw with the tensor product of vector spaces, the dual space construction turns out to be a functor. In fact, it’s a contravariant functor. That is, if we have a linear transformation we get a linear transformation . As usual, we ask what this looks like for matrices.

First, how do we define the dual transformation? It turns out this is the contravariant functor represented by . That is, if is a linear functional, we define . In terms of the action on vectors,

Now let’s assume that and are finite-dimensional, and pick bases and for and , respectively. Then the linear transformation has matrix coefficients . We also get the dual bases of and of .

Given a basic linear functional on , we want to write in terms of the . So let’s evaluate it on a generic basis vector and see what we get. The formula above shows us that

In other words, we can write . The same matrix works, but we use its indices differently.

In general, given a linear functional with coefficients we find the coefficients of as . The value becomes . Notice that the summation convention tells us this must be a scalar (as we expect) because there are no unpaired indices. Also notice that because we can use the same matrix for two different transformations we seem to have an ambiguity: is the lower index running over a basis for or one for ? Luckily, since every basis gives rise to a dual basis, we don’t need to care. Both spaces have the same dimension anyhow.

[...] the linear functional on into one on . But this is just the dual transformation ! Then we can evaluate this on the column vector to get the same result: [...]

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[...] the way we defined the dual transformation was such that we can instead apply the dual to the linear functional , and [...]

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