# 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.

Posted by John Armstrong | Algebra, Linear Algebra | 26 Comments

## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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