18 - Adjoints of Maps

Adjoint of map

Let $V$ be a inner product space over $\mathbb{K}=\mathbb{R},\mathbb{C}$

Given a linear map $T:V\to V$ then
Linear map $T^{\ast }:V\to V$ is the adjoint of $T$ if

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

Adjoint Map is Unique lemma

Given a linear map $T:V\to V$ then

$$\text{if }{T}^{\ast }\text{ exists it is unique}$$

Proof

Let $\tilde{T}$ be another map satisfying the adjoint definition then
For all $v,w\in V$

$$\begin{array}{rl}⟨T^{\ast }(v)-\tilde{T}(v),w⟩& =⟨T^{\ast }(v),w⟩-⟨\tilde{T}(v),w⟩\\ & =⟨v,T(w)⟩-⟨v,T(w)⟩\\ & =0\end{array}$$

As $⟨⟩$ is non-degenerate then for all $v\in V$

$$T^{\ast }(v)-\tilde{T}(v)=0$$

Hence $T^{\ast }=\tilde{T}$

Matrix Representation of the Adjoint

Let $T:V\to V$ be linear and let $\mathcal{B}=\{e_1,\cdots ,e_n\}$ be an orthonormal basis for $V$ then

$${}_{\mathcal{B}}[T^{\ast }{]}_{\mathcal{B}}={\overline{{}_{\mathcal{B}}[T{]}_{\mathcal{B}}}}^T$$

Proof

Let $A={}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ then

$$a_{ij}=⟨e_i,T(e_j)⟩$$

So let $B={}_{\mathcal{B}}[T^{\ast }{]}_{\mathcal{B}}$ then

$$b_{ij}=⟨e_i,T^{\ast }(e_j)⟩=\overline{⟨T^{\ast }(e_j),e_i⟩}=\overline{⟨e_j,T^{\ast }(e_i)⟩}=\overline{a_{ji}}$$

Hence $B={\overline{A}}^T$

Properties of Adjoint Maps

Let $S,T:V\to V$ be linear
Let $V$ be finite dimensional

Let $\lambda \in \mathbb{K}$ then

$$(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$$

$$(\lambda T{)}^{\ast }=\overline{\lambda }{T}^{\ast }$$

$$(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$$

$$(T^{\ast }{)}^{\ast }=T$$

5. If $m_T$ is the minimal polynomial of $T$ then

$$m_{T^{\ast }}=\overline{m_T}$$

Self-Adjoint Map

Let $T:V\to V$ be a linear map then

$$T\text{ is self-adjoint}\text{ if }T=T^{\ast }$$

Eigenvalues of Self-Adjoint Linear Operators lemma

If $\lambda$ is an eigenvalue of a self-adjoint linear operator then

$$\lambda \in \mathbb{R}$$

Proof

Assume $w\ne 0$ and $T(w)=\lambda w$ for some $\lambda \in \mathbb{C}$ then

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

As $⟨w,w⟩\ne 0$ then $\lambda =\overline{\lambda }$
Hence

$$\lambda \in \mathbb{R}$$

Orthogonal Complements of Invariant Subspaces lemma

Let $T$ be a self-adjoint map
Let $U\subseteq V$ and $U$ is $T$-invariant then

$$U^{\perp }\text{ is }T-\text{invariant}$$

Proof

Let $w\in U^{\perp }$ then for all $u\in U$

$$⟨u,T(w)⟩=⟨T^{\ast }(u),w⟩=⟨T(u),w⟩=0$$

as $T(u)\in U$ and $w\in U^{\perp }$ hence $T(w)\in U^{\perp }$