Unitary Matrix

Definition

${\mathbf{U}}^{†}\mathbf{U}=\mathbf{U}{\mathbf{U}}^{†}=\mathbf{I}$ and ${\mathbf{U}}^{†}={\mathbf{U}}^{-1}$

Facts

$⟨\mathbf{Ux},\mathbf{Uy}⟩=⟨\mathbf{x},\mathbf{y}⟩$ Given two complex vectors $\mathbf{x},\mathbf{y}$, multiplication by $\mathbf{U}$ preserves their Inner product.

$||\mathbf{x}||=||\mathbf{Ux}||$ Given a complex vector $\mathbf{x}$, multiplication by $\mathbf{U}$ preserves its Norm.

$|{\lambda }_i|=1$ Every Eigenvalue of $\mathbf{U}$ has unit length. Therefore, $|U|=\pm 1$

Eigenvectors of $\mathbf{U}$ are orthogonal

Proof For the Eigendecomposition $\mathbf{U}{\mathbf{x}}_1={\lambda }_1{\mathbf{x}}_1,\mathbf{U}{\mathbf{x}}_2={\lambda }_2{\mathbf{x}}_2$ where ${\lambda }_1\ne {\lambda }_2$, ${\mathbf{x}}_1^{†}{\mathbf{x}}_2=(\mathbf{U}{\mathbf{x}}_1{)}^{†}(\mathbf{U}{\mathbf{x}}_2)=({\lambda }_1{\mathbf{x}}_1{)}^{†}({\lambda }_2{\mathbf{x}}_2)={\bar{\lambda }}_1{\lambda }_2{\mathbf{x}}_1^{†}{\mathbf{x}}_2$. So, $(1-{\bar{\lambda }}_1{\lambda }_2){\mathbf{x}}_1^{†}{\mathbf{x}}_2=0$. Since ${\bar{\lambda }}_1{\lambda }_2=1$ only where ${\lambda }_1={\lambda }_2$, by the assumption ${\mathbf{x}}_1^{†}{\mathbf{x}}_2=0$.

$\mathrm{tr}(U^{⊺}AU)=\mathrm{tr}(A)$ $|U^{⊺}AU|=|A|$