Linear Functionals

We will now look at a new type of function that takes vectors $v \in V$ and maps to scalars in $\mathbb{F}$.

Definition: A Linear Functional is a linear map $\varphi : V \to \mathbb{F}$ from the vector space $V$ to the field of scalars $\mathbb{F}$.

For example, consider the vector space $\mathbb{R}^4$ and the field $\mathbb{R}$. Then for any vector $x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4$ we can define a linear functional on $\mathbb{R}^4$ for $a, b, c, d \in \mathbb{R}$ by:

(1)

\begin{align} \quad \varphi (x) = \varphi (x_1, x_2, x_3, x_4) = ax_1 + bx_2 + cx_3 + dx_4 \end{align}

(Note that $ax_1 + b_2 + cx_3 + dx_4) \in \mathbb{R}$). For a more precise example, let $a = b = c = d = 1$. Then one such linear functional from $\mathbb{R}^4$ to $\mathbb{R}$ is thus given by:

(2)

\begin{align} \quad \varphi(x) = \varphi (x_1, x_2, x_3, x_4) = x_1 + x_2 + x_3 + x_4 \end{align}

Suppose that $V$ is an inner product space, and suppose that we fix a vector $v \in V$, then we can define a linear function that takes every vector $u \in V$ to $<u, v>$ as:

(3)

\begin{align} \quad \varphi (u) = <u, v> \end{align}

In fact, every linear function on the vector space $V$ will be in this form as the following theorem establishes.

Theorem 1: Let $V$ be a finite-dimensional inner product space and let $\varphi$ be a linear functional on $V$. Then there exists a unique vector $v \in V$ such that for every $u \in V$ we have that $\varphi (u) = <u, v>$.

• Proof: Since $V$ is a finite-dimensional inner product space, then let $\{ e_1, e_2, ..., e_n \}$ be an orthonormal basis of $V$. Then for any vector $u \in U$ we have that $u = <u, e_1>e_1 + <u, e_2>e_2 + ... + <u, e_n>e_n$. Since $\varphi$ is a linear functional (and hence a linear map), we have that:

(4)

\begin{align} \quad \varphi(u) = \varphi(<u, e_1>e_1 + <u, e_2>e_2 + ... + <u, e_n>e_n>) \\ \quad \varphi(u) = <u, e_1>\varphi(e_1) + <u, e_2>\varphi(e_2) + ... + <u, e_n> \varphi(e_n) \\ \end{align}

• Note that $\varphi(e_1), \varphi(e_2), ..., \varphi(e_n) \in \mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$), and we we can use the property of conjugate homogeneity in the second slot of the inner product to get:

(5)

\begin{align} \quad \varphi(u) = <u, \overline{\varphi(e_1)}e_1> + <u, \overline{\varphi(e_2)e_2} + ... + <u, \overline{\varphi(e_n)e_n} \\ \quad \varphi(u) = <u, \overline{\varphi(e_1)}e_1 + \overline{\varphi(e_2)}e_2 + ... + \overline{\varphi(e_n)}e_n> \end{align}

• So let $v = \overline{\varphi(e_1)}e_1 + \overline{\varphi(e_2)}e_2 + ... + \overline{\varphi(e_n)}e_n$. Thus for every $u \in V$ we have that $\varphi(u) = <u, v>$.

• We now need to show that this vector $v$ is unique. Suppose that for all $u \in V$ there exists two vectors $v, v' \in V$ such that $\varphi(u) = <u, v> = <u, v'>$. Then we have that for every $u \in V$:

(6)

\begin{align} \quad 0 = <u, v> - <u, v'> \\ \quad 0 = <u, v - v'> \end{align}

• Now suppose that $u = v - v'$. Then it must hold that $<v - v', v - v'> = 0$. By the definiteness property of an inner product we have that this implies $v - v' = 0$, i.e, $v = v'$. So indeed, the vector $v \in V$ for which every $u \in U$ we have $\varphi(u) = <u, v>$ is unique. $\blacksquare$