# The Unapologetic Mathematician

## 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

$\left[T^*(\phi^l)\right](e_i)=\phi^l(T(e_i))=\phi^l(t_i^kf_k)=t_i^k\delta_k^l=t_i^l$

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.

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Posted by John Armstrong | Algebra, Linear Algebra

## 2 Comments	»

1. [...] 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: [...]

Pingback by | May 30, 2008 | Reply

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

Pingback by | May 22, 2009 | Reply

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