Adjoints Exist for Finite Dimensional Space

Definition

For Linear Operator $T$ on finite dimensional Inner Product Space, there exists unique Linear Operator $T^{\ast }$ on $V$ such that $\forall \alpha ,\beta \in V$, $⟨T\alpha |\beta ⟩=⟨\alpha |T^{\ast }\beta ⟩$

Proof

1. Let $\beta$ be any vector $\in V$, $T\in \mathcal{L}(V)$
2. Let $f$ be the Linear Functional mapping $\alpha$ to $⟨T\alpha |\beta ⟩$
3. Now, $f(\alpha )=⟨T\alpha |\beta ⟩$
4. By construction, $f$ is a linear function
5. Thus, $\exists {\beta }^{\prime }\in V$ such that $f(\alpha )=⟨\alpha |{\beta }^{\prime }⟩=⟨T\alpha |\beta ⟩,\forall \alpha \in V$
6. Let $T^{\ast }$ map $\beta$ to ${\beta }^{\prime }$
7. Then, $T^{\ast }(\beta )={\beta }^{\prime }$
8. Thus, the inner product $⟨\alpha |{\beta }^{\prime }⟩=⟨\alpha |T^{\ast }(\beta )⟩=⟨T\alpha |\beta ⟩,\forall \alpha \in V$

Proving $T^{\ast }$ is linear and unique

1. $⟨\alpha |T_1^{\ast }\beta ⟩=⟨\alpha |T_2^{\ast }\beta ⟩,\forall \alpha ,\beta \in V$
2. Now, we want to show $T_1^{\ast }\beta =T_2^{\ast }\beta ,\forall \beta \in V$
3. Note $⟹⟨\alpha |T_1^{\ast }\beta ⟩-⟨\alpha |T_2^{\ast }\beta ⟩=0,\forall \beta \in V$
4. $⟹⟨\alpha |T_1^{\ast }\beta -T_2^{\ast }\beta ⟩=0,\forall \beta \in V$
5. $⟹T_1^{\ast }\beta -T_2^{\ast }\beta =0,\forall \beta \in V$
6. $⟺T_1^{\ast }\beta =T_2^{\ast }\beta ,\forall \beta \in V$