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

\begin{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.
\end{proof}
There's many other things we could prove, but we'll save you on time.

Properties of the adjoint

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

\begin{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 $I = I^{*}$ .
\end{proof}
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 $w \in W$ .

\begin{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.
\end{proof}
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.

\begin{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.
\end{proof}

Matrix Representation of the Adjoint

null and range of T∗

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