Suggest edit — linear functional

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---
title: "linear functional"
type: concept
related: [Orthonormal Basis, Linear Map, Linear Functional, Inner Product, Finite Dimensional Vector Space]
source: https://www.jemoka.com/posts/kbhlinear_functional/
confidence: high
status: active
---

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 \(e_1, &hellip;e_{n}\) be an orthonormal basis of \(V\). Recall there&rsquo;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) &amp;= \varphi(\langle v, e_1 \rangle e_1 + \dots \langle v, e_{n} \rangle e_{n}) \\ &amp;= \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) &amp;= \varphi(\langle v, e_1 \rangle e_1 + \dots \langle v, e_{n} \rangle e_{n}) \\ &amp;= \langle v, e_1 \rangle \varphi(e_1) + \dots + \langle v, e_n \rangle \varphi(e_n) \\ &amp;= \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. \(\blacksquare\)