Lecture 8 - Linear Functionals

Let $V$ be an finite-dimensional inner product space. If $v \in V$ is fixed, then;

⟨ \cdot, v ⟩ \in L (V, F) = V^{'}

It's a fact then that every linear functional is equal to $⟨ \cdot, v ⟩$ for some vector $v$ . If you give me a linear function $φ$ then I can give you a vector $v$ such that:

φ (u) = ⟨ u, v ⟩

This is the Risz Representation Theorem:

Riesz Representation Theorem

Suppose $V$ is finite-dimensional and $φ$ is a linear functional on $V$ . Then there is a unique vector $u \in V$ such that:
$φ (v) = ⟨ v, u ⟩$
for every $v \in V$ .

The proof is seen via Chapter 6 (cont.) - Finishing Inner Product Spaces#^dff70d.

Here, the association is the idea that $φ$ is a map from $V$ to $V^{'}$ :

Here: $φ_{V} : V \to V^{'}$ , where:

φ_{V} (v_{1} + v_{2}) = ⟨ \cdot, v_{1} ⟩ + ⟨ \cdot, v_{2} ⟩ = φ_{V} (v_{1}) + φ_{V} (v_{2})

So additivity holds. Homogeneity is and interesting story:

φ_{V} (α v) = ⟨ \cdot, α v ⟩ = \overset{―}{α} φ_{V} (v)

Notice that the $α$ is the conjugate, hence $φ_{V}$ is a conjugate linear map. If you are working with a real vector space, then $φ_{V}$ is an isomorphism (so $V ≃ V^{'}$ ), but in the complex vector spaces then it's not necessarily the case.

More in Depth Look at Finite-Dimensional Inner Product Spaces

We now on will restrict entirely to finite-dimensional inner product spaces. Every vector space you see from here on will be finite-dimensional, even if not explicitly stated.

Recall the Riesz Representation Theorem:

Riesz Representation Theorem

Suppose $V$ is finite-dimensional and $φ$ is a linear functional on $V$ . Then there is a unique vector $u \in V$ such that:
$φ (v) = ⟨ v, u ⟩$
for every $v \in V$ .

So if you give me $T \in L (V, W)$ , and some vector $w \in W$ ,

We're going to consider $φ \in V^{'}$ defined:

φ (v) = ⟨ T v, w ⟩

this is a linear functional on $V$ , so then by the Riesz, then:

φ (v) = ⟨ T v, w ⟩ = ⟨ v, \underset{some vector in V}{\underset{⏟}{_}} ⟩

it seems that this will depend on $T$ and $w$ . Let's, for the sake of time, call it $T^{*} (w)$ . We'll show that $T^{*}$ is a linear map, but we'll get to that.

adjoint

The adjoint of $T \in L (V, W)$ is the function $T^{*} : W \to V$ satisffying:

⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩

for all $v \in V, w \in W$ .

Let's see what this adjoint look like in an example. Define $T : R^{2} \to R^{4}$ be defined by:

T (x_{1}, x_{2}) = (x_{1}, x_{2}, x_{1} + x_{2}, x_{1} - x_{2})

It's not hard to show that this $T \in L (R^{2}, R^{4})$ . In fact, we could determine the matrix:

M (T) = [\begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \\ 1 & - 1 \end{matrix}]

Let's find $T^{*}$ . We need the inner product:

\begin{aligned} ⟨ (x_{1}, x_{2}), T^{*} (y_{1}, y_{2}, y_{3}, y_{4}) ⟩ & = ⟨ T (x_{1}, x_{2}), (y_{1}, y_{2}, y_{3}, y_{4}) ⟩ \\ = ⟨ (x_{1}, x_{2}, x_{1} + x_{2}, x_{1} - x_{2}), (y_{1}, y_{2}, y_{3}, y_{4}) ⟩ \\ = x_{1} y_{1} + x_{2} y_{2} + (x_{1} + x_{2}) y_{3} + (x_{1} - x_{2}) y_{4} \\ = ⟨ (x_{1}, x_{2}), (y_{1} + y_{3} + y_{4}, y_{2} + y_{3} - y_{4}) ⟩ \end{aligned}

equating the RHS of the brakets gives that $T^{*} (y_{1}, y_{2}, y_{3}, y_{4}) = (y_{1} + y_{3} + y_{4}, y_{2} + y_{3} - y_{4})$ .

What if we wanted to see $M (T^{*})$ :

\begin{aligned} T^{*} (1, 0, 0, 0) & = (1, 0) \\ T^{*} (0, 1, 0, 0) & = (0, 1) \\ T^{*} (0, 0, 1, 0) & = (1, 1) \\ T^{*} (0, 0, 0, 1) & = (1, - 1) \end{aligned}

Thus:

M (T^{*}) = [\begin{matrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & - 1 \end{matrix}] = M (T)^{T}

the transpose is back at it again! But in general, the matrix that represents the adjoint is the conjugate transpose of $M (T)$ . Tomorrow, we'll show that this is the general case.