A linear map to numbers. Its very powerful because any linear functional can be represented as an inner product using Riesz Representation Theorem

## constituents

- vector space \(V\)
- a linear map \(\varphi \in \mathcal{L}(V, \mathbb{F})\)

## requirements

\(\varphi\) is called a linear functional on \(V\) if \(\varphi: V \to \mathbb{F}\). That is, it maps elements of \(V\) to scalars. For instance, every inner product is a Linear Map to scalars and hence a linear functional.

## additional information

### Riesz Representation Theorem

Suppose \(V\) is finite-dimensional, and \(\varphi\) is a linear functional on \(V\); then, there exists an **unique** \(u \in V\) such that:

\begin{equation} \varphi(v) = \langle v,u \rangle \end{equation}

\(\forall v \in V\). Kinda a mindblowing fact.

Proof:

Every Inner Product Space has an orthonormal basis; let \(e_1, …e_{n}\) be an orthonormal basis of \(V\). Recall there’s a specific way of writing a vector as a linear combination of orthonormal basis, that WLOG \(v \in V\):

\begin{equation} v = \langle v, e_1 \rangle e_1 + \dots \langle v, e_{n} \rangle e_{n} \end{equation}

Now:

\begin{equation} \varphi(v) = \varphi(\langle v, e_1 \rangle e_1 + \dots \langle v, e_{n} \rangle e_{n}) \end{equation}

Given homogenity and addtivity, we then have:

\begin{align} \varphi(v) &= \varphi(\langle v, e_1 \rangle e_1 + \dots \langle v, e_{n} \rangle e_{n}) \\ &= \langle v, e_1 \rangle \varphi(e_1) + \dots + \langle v, e_n \rangle \varphi(e_n) \end{align}

Now, shoving \(\varphi\) into the second slot (remember we have conjugate homogenity on the secon slot), and adding it all together (as inner products are additive in both slots):

\begin{align} \varphi(v) &= \varphi(\langle v, e_1 \rangle e_1 + \dots \langle v, e_{n} \rangle e_{n}) \\ &= \langle v, e_1 \rangle \varphi(e_1) + \dots + \langle v, e_n \rangle \varphi(e_n) \\ &= \langle v, \overline{\varphi(e_1)} e_1 + \dots + \overline{\varphi(e_n)}e_n \rangle \end{align}

You will note now that the second slot to this inner product is **v-independent!** So as long as we know the orthonormal basis we can encode \(\varphi\) with:

\begin{equation} u = \overline{\varphi(e_1)} e_1 + \dots + \overline{\varphi(e_n)}e_n \end{equation}

and:

\begin{equation} \varphi(v) = \langle v, u \rangle \end{equation}

Now, to show uniqueness, we probably do the same damned thing we have a million times:

Suppose:

\begin{equation} \varphi(v) = \langle v,u_1 \rangle = \langle v,u_{2} \rangle \end{equation}

holds for all \(v \in V\), as required by the theorem.

This means that:

\begin{equation} \langle v, u_1-u_2 \rangle = 0 \end{equation}

For every \(v \in V\). Let \(v = u_1-u_2\). Now by definiteness we have \(u_1-u_2=0\) meaning \(u_1=u_2\) as desired. \(\blacksquare\)