linear functional — Jemoka Knowledge Base

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:

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