Adjoints

Adjoint

Proposition

Let T\colon \H\to\H be a bounded linear map on a separable Hilbert space \H. Then there exists a unique bounded linear map T^{*}\colon \H\to\H such that $⟨Tf,g⟩=⟨f,T^{\ast }g⟩$ for all f,g\in\H. We call $T^{\ast }$ the adjoint of $T$.

Proposition

The operator norm of the adjoint is equal to the operator norm of the original operator, i.e. $\Vert T^{\ast }\Vert =\Vert T\Vert$.

Infinite Diagonal Matrix

Diagonalised Linear Operator

Suppose \H is a separable Hilbert space with orthonormal basis $\{e_i{\}}_{i=1}^{\infty }$. A linear operator T\colon \H\to\H is diagonalised if there exists a sequence of scalars $\{{\lambda }_i{\}}_{i=1}^{\infty }\subset \text{C}$ such that $T(e_i)={\lambda }_i{e}_i,\text{for all }i.$

Proposition

Suppose T\colon \H\to\H is a diagonalised linear operator on a separable Hilbert space \H with orthonormal basis $\{e_i{\}}_{i=1}^{\infty }$. Then the following holds:

• $\Vert T\Vert ={\sup }_k|{\lambda }_k|$,
• $T^{\ast }$ is also diagonalised with ${\bar{\lambda }}_k$ as the diagonal entries. Hence, $T=T^{\ast }$ if and only if ${\lambda }_k\in \mathbb{R}$ for all $k$.
• $T$ is unitary if and only if $|{\lambda }_k|=1$ for all $k$.
• $T$ is an orthogonal projection if and only if ${\lambda }_k\in \{0,1\}$ for all $k$.