The Unapologetic Mathematician

Mathematics for the interested outsider

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 | 26 Comments

   

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