The Unapologetic Mathematician

Mathematics for the interested outsider

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 T:U\rightarrow V we get a linear transformation T^*:V^*\rightarrow U^*. 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 \mathbb{F}. That is, if \mu:V\rightarrow\mathbb{F} is a linear functional, we define T^*(\mu)=\mu\circ T:U\rightarrow\mathbb{F}. In terms of the action on vectors, \left[T^*(\mu)\right](v)=\mu(T(v))

Now let’s assume that U and V are finite-dimensional, and pick bases \left\{e_i\right\} and \left\{f_k\right\} for U and V, respectively. Then the linear transformation T has matrix coefficients t_i^k. We also get the dual bases \left\{\epsilon^j\right\} of U^* and \left\{\phi^l\right\} of V^*.

Given a basic linear functional \phi^l on V, we want to write T^*(\phi^l) in terms of the \epsilon^j. So let’s evaluate it on a generic basis vector e_i and see what we get. The formula above shows us that


In other words, we can write T^*(\phi^l)=t_j^l\epsilon^j. The same matrix works, but we use its indices differently.

In general, given a linear functional \mu with coefficients \mu_l we find the coefficients of T^*(\mu) as t_j^l\mu_l. The value \left[T^*(\mu)\right](v)=\mu(T(u)) becomes \mu_lt_i^lu^i. 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 U or one for U^*? Luckily, since every basis gives rise to a dual basis, we don’t need to care. Both spaces have the same dimension anyhow.

May 28, 2008 Posted by | Algebra, Linear Algebra | 2 Comments