Lecture 9 - Finishing Linear Functional Via Inner Products

Recall:

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

We said a few facts about $T^{*}$ , namely:

Lemma

$T^{*}$ is linear

Proof
Given two vectors $w_{1}, w_{2} \in W$ , we want to show that $T^{*} (w_{1} + w_{2}) = T^{*} w_{1} + T^{*} w_{2}$ . But it's hard to start with the left hand side, so to prove it we have to introduce the inner product. We need a handy lemma:

Lemma

$⟨ v, u_{1} ⟩ = ⟨ v, u_{2} ⟩$ for all $v \in V$ is equivalent to $u_{1} = u_{2}$ .

As such, we just need to show that the inner product of the vectors against the two vectors above are equal. Let's do that. Let $v \in V$ :

\begin{aligned} ⟨ v, T^{*} (w_{1} + w_{2}) ⟩ & = ⟨ T v, w_{1} + w_{2} ⟩ \\ = ⟨ T v, w_{1} ⟩ + ⟨ T v, w_{2} ⟩ \\ = ⟨ v, T^{*} w_{1} ⟩ + ⟨ v, T^{*} w_{2} ⟩ \end{aligned}

Thus $T^{*} (w_{1} + w_{2}) = T^{*} (w_{1}) + T^{*} (w_{2})$ using Lecture 9 - Finishing Linear Functional Via Inner Products#^9c0ab3, so then $T^{*}$ is additive. For homogeneity, let $λ \in F$ be arbitrary. Then:

\begin{aligned} ⟨ v, T^{*} (λ w_{1}) ⟩ & = ⟨ T v, λ w_{1} ⟩ \\ = \overset{―}{λ} ⟨ T v, w_{1} ⟩ \\ = \overset{―}{λ} ⟨ v, T^{*} w_{1} ⟩ \\ = ⟨ v, λ T^{*} w_{1} ⟩ \end{aligned}

Thus, in a similar way, then $λ T^{*} (w_{1}) = T^{*} (λ w_{1})$ , showing homogeneity.
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There's many other things we could prove, but we'll save you on time.

Properties of the adjoint

$(S + T)^{*} = S^{*} + T^{*}$
$(λ T)^{*} = \overset{―}{λ} T^{*}$
$(T^{*})^{*} = T$
$I^{*} = I$
$(S T)^{*} = T^{*} S^{*}$

Proof
Let's prove (c) for fun. Let $v \in V, w \in W$ . Looking at the inner product:

\begin{aligned} ⟨ w, (T^{*})^{*} v ⟩ & = ⟨ T^{*} w, v ⟩ \\ = \overset{―}{⟨ v, T^{*} w ⟩} \\ = \overset{―}{⟨ T v, w ⟩} \\ = ⟨ w, T v ⟩ \end{aligned}

Thus by uniqueness lemma, then $T = (T^{*})^{*}$ .

Another one, let's do (d). Consider where $I \in L (V)$ :

\begin{aligned} ⟨ v, I^{*} v_{1} ⟩ & = ⟨ I v, v_{1} ⟩ \\ = ⟨ v, v_{1} ⟩ \\ = ⟨ v, I v_{1} ⟩ \end{aligned}

So then $I = I^{*}$ .
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Recall that the adjoint of the matrix was the conjugate of the transpose of $M (T)$ . See Lecture 8 - Linear Functionals#^a25838 for more info.

here we have:

$φ_{V} : V \to V^{'}$ defined by $φ_{V} (v) = ⟨ \cdot, v ⟩$ . Similar for $φ_{W}$ .
Here $T \in L (V, W)$ and $T^{*} \in L (W, V)$ .
Thus, $T^{'} \in L (W^{'}, V^{'})$ , taking linear functionals from $W^{'}$ and creating a new one in $V^{'}$ .

We claim that $T^{*} = φ_{V}^{- 1} \circ T^{'} \circ φ_{W}$ , using the picture from above.

Theorem

$T^{*} = φ_{V}^{- 1} \circ T^{'} \circ φ_{W}$ .

We'll use the fact that $⟨ v, T^{*} w ⟩ = ⟨ T v, w ⟩$ for all $v \in V$ and $w \in W$ .

Proof

\begin{aligned} \underset{⟨ \cdot, T^{*} w ⟩}{\underset{⏟}{⟨ v, T^{*} w ⟩}} & = \underset{⟨ \cdot, w ⟩}{\underset{⏟}{⟨ T v, w ⟩}} \\ φ_{V} (T^{*} w) v & = φ_{W} (w) T v \\ = (φ_{W} (w) \circ T) v \end{aligned}

So then $φ_{V} (T^{*} w) = φ_{W} (w) \circ T$ . But this is the dual map on the RHS, which is:

\begin{aligned} φ_{V} (T^{*} w) & = φ_{W} (w) \circ T \\ = T^{'} (φ_{W}) \\ T^{*} (w) & = φ_{V}^{- 1} \circ T^{'} \circ φ_{W} (w) \end{aligned}

Notice that $\exists φ_{V}^{- 1}$ since all linear functionals are injective, mainly because these are finite-dimensions. As such, then the theorem is proved.
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Notice specifically that this is a way to show that $T^{*}$ is linear, since all $φ_{W}, T^{'}, φ_{V}$ are linear. Further, $φ_{V}, φ_{W}$ have conjugate homogeneity, so that actually makes $T^{*}$ a fully linear transformation.

Matrix Representation of the Adjoint

Let's do the proof.

We'll use the orthonormal basis $β_{V} = {e_{1}, . . ., e_{n}}$ for $V$ , and $β_{W} = {f_{1}, . . ., f_{m}}$ for $W$ . So then:

\begin{aligned} M (T^{*}, β_{W}, β_{V})_{i j} & = \underset{⟨ v, e_{1} ⟩ e_{1} + \dots + ⟨ v, e_{n} ⟩ e_{n}}{\underset{⏟}{⟨ T^{*} (f_{j}), e_{i} ⟩}} e_{i} \\ = \overset{―}{⟨ T e_{i}, f_{j} ⟩} \\ = \overset{―}{M (T, β_{V}, β_{W})_{j i}} \end{aligned}

Thus $M (T^{*})$ is the conjugate transpose of $M (T)$ . $A^{*}$ is the conjugate transpose of $A$ .

$null$ and $range$ of $T^{*}$

$null (T^{*}) = (range (T))^{⊥}$
$range (T^{*}) = (null (T))^{⊥}$
$null (T) = (range (T^{*}))^{⊥}$
$range (T) = (null (T^{*}))^{⊥}$

Note that (a,d) are the same thing, and similarly (b,c) as well. Further (a,c) are equivalent, so then we only really need to show one of them.

Proof
We prove (a). Suppose $w \in null (T^{*})$ so then $T^{*} (w) = {\vec{0}}_{V}$ . So then:

\begin{aligned} ⟨ v, T^{*} w ⟩ & = 0 \\ ⟨ T v, w ⟩ & = 0 \end{aligned}

for all $v \in V$ , so then we must have it that $w \in (range (T))^{⊥}$ , as otherwise the inner product is not 0.
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Matrix Representation of the Adjoint

null and range of T∗

$null$ and $range$ of $T^{*}$