# The Unapologetic Mathematician

## Dual Spaces

Another thing vector spaces come with is duals. That is, given a vector space $V$ we have the dual vector space $V^*=\hom(V,\mathbb{F})$ of “linear functionals” on $V$ — linear functions from $V$ to the base field $\mathbb{F}$. Again we ask how this looks in terms of bases.

So let’s take a finite-dimensional vector space $V$ with basis $\left\{e_i\right\}$, and consider some linear functional $\mu\in V^*$. Like any linear function, we can write down matrix coefficients $\mu_i=\mu(e_i)$. Notice that since our target space (the base field $\mathbb{F}$) is only one-dimensional, we don’t need another index to count its basis.

Now let’s consider a specially-crafted linear functional. We can define one however we like on the basis vectors $e_i$ and then let linearity handle the rest. So let’s say our functional takes the value ${1}$ on $e_1$ and the value ${0}$ on every other basis element. We’ll call this linear functional $\epsilon^1$. Notice that on any vector we have

$\epsilon^1(v)=\epsilon^1(v^ie_i)=v^i\epsilon^1(e_i)=v^1$

so it returns the coefficient of $e_1$. There’s nothing special about $e_1$ here, though. We can define functionals $\epsilon^j$ by setting $\epsilon^j(e_i)=\delta_i^j$. This is the “Kronecker delta”, and it has the value ${1}$ when its two indices match, and ${0}$ when they don’t.

Now given a linear functional $\mu$ with matrix coefficients $\mu_j$, let’s write out a new linear functional $\mu_j\epsilon^j$. What does this do to basis elements?

$\mu_j\epsilon^j(e_i)=\mu_j\delta_i^j=\mu_i$

so this new transformation has exactly the same matrix as $\mu$ does. It must be the same transformation! So any linear functional can be written uniquely as a linear combination of the $\epsilon^j$, and thus they form a basis for the dual space. We call $\left\{\epsilon^j\right\}$ the “dual basis” to $\left\{e_i\right\}$.

Now if we take a generic linear functional $\mu$ and evaluate it on a generic vector $v$ we find

$\mu(v)=\mu_j\epsilon^j(v^ie_i)=\mu_jv^i\epsilon^j(e_i)=\mu_jv^i\delta_i^j=\mu_iv^i$

Once we pick a basis for $V$ we immediately get a basis for $V^*$, and evaluation of a linear functional on a vector looks neat in terms of these bases.

May 27, 2008 Posted by | Algebra, Linear Algebra | 30 Comments