26 - Existence and Linearity of an Adjoint Map

Existence of an Adjoint Map

Let $T:V\to V$ be linear where $V$ is finite dimensional then

$$\text{adjoint }{T}^{\ast }\text{ exists and is linear }$$

Proof

Let $v\in V$ and consider map $V\to \mathbb{K}$ given by

$$w\mapsto ⟨v,T(w)⟩$$

then

$$⟨v,T(\cdot )⟩$$

is a linear functional as $T$ is linear so $⟨⟩$ is linear in second coordinate

As $V$ is finite dimensional then $\varphi :V\to V^{\prime }$ given by $\varphi (u)=⟨u,\cdot ⟩$
is a $R$-linear isomorphism and in particular a surjective map

So there exists $u\in V$ such that

$$⟨v,T(\cdot )⟩=⟨u,\cdot ⟩$$

By defining $T^{\ast }(v):=u$ then

$$⟨v,T(\cdot )⟩=⟨T^{\ast }(v),\cdot ⟩$$

In other words then

$$⟨v,T(w)⟩=⟨T^{\ast }(v),w⟩\text{ for all }w\in V$$

To see that $T^{\ast }$ is linear, note that for all $v_1,v_2,w\in V$ and $\lambda \in \mathbb{K}$ then

$$\begin{array}{rl}⟨T^{\ast }(v_1+\lambda v_2),w⟩& =⟨v_1+\lambda v_2,T(w)⟩\\ & =⟨v_1,T(w)⟩+\overline{\lambda }⟨v_2,T(w)⟩\\ & =⟨T^{\ast }(v_1),w⟩+\overline{\lambda }⟨T^{\ast }(v_2),w⟩\\ & =⟨T^{\ast }(v_1)+\lambda T^{\ast }(v_2),w⟩\end{array}$$

As $⟨⟩$ is non-degenerate then

$$T^{\ast }(v_1+v_2)=T^{\ast }(v_1)+\lambda T^{\ast }(v_2)$$