Normal matrices and diagonalizability

Theorem: The product of two unitary matrices is unitary.

Proof: Let ${\bf U}$ and ${\bf V}$ be unitary, i.e., ${\bf U}^*={\bf U}^{-1}$ and ${\bf V}^*={\bf V}^{-1}$, then ${\bf U}{\bf V}$ is unitary:

$$({\bf U}{\bf V})^*={\bf V}^*{\bf U}^*={\bf V}^{-1}{\bf U}^{-1}=({\bf U}{\bf V})^{-1}$$

Theorem: Two square matrices ${\bf A}$ and ${\bf B}$ are simultaneously diagonalizable if and only if they commute.

Proof (reference)

• Let ${\bf A}$ and ${\bf B}$ be simultageously diagonalizable by ${\bf R}$
  
  $${\bf R}^{-1}{\bf A}{\bf R}={\bf\Lambda}_A, \;\;\;\;\;\;\;\; {\bf R}^{-1}{\bf B}{\bf R}={\bf\Lambda}_B$$
  
  
  then
  
  $${\bf A}{\bf B} =({\bf R}{\bf\Lambda}_A{\bf R}^{-1})({\bf R... ... R}^{-1})({\bf R}{\bf\Lambda}_A{\bf R}^{-1}) ={\bf B}{\bf A}$$
  
  

• Let ${\bf A}$ and ${\bf B}$ commute, i.e., ${\bf A}{\bf B}={\bf B}{\bf A}$. Assuming ${\bf u}$ is an eigenvector of ${\bf A}$ corresponding to eigenvalue $\lambda$, i.e., ${\bf A}{\bf u}=\lambda{\bf u}$, then
  
  $${\bf A}{\bf B}{\bf u}={\bf B}({\bf A}{\bf u})=\lambda{\bf B}{\bf u}$$
  
  
  we see that ${\bf B}{\bf u}$ is also an eigenvector of ${\bf A}$ corresponding to the same eigenvalue $\lambda$, i.e., ${\bf B}{\bf u}$ must be a scaled version of ${\bf u}$ (in the same 1-D space): ${\bf B}{\bf u}=\gamma{\bf u}$, i.e., ${\bf u}$ is also an eigenvector of ${\bf B}$.

Theorem: A matrix is normal if and only if it is unitarily diagonalizable.

Proof (reference)

• If ${\bf A}$ is unitarily diagonalizable:
  
  $${\bf A}{\bf U}={\bf U}{\bf\Lambda},\;\;\;\;\;\;\; {\bf U}^... ...ambda}, \;\;\;\;\;\;\; {\bf A}={\bf U}{\bf\Lambda}{\bf U}^*$$
  
  
  where ${\bf U}^*={\bf U}^{-1}$ is unitary and ${\bf\Lambda}$ is a diagonal matrix satisfying ${\bf\Lambda}^*{\bf\Lambda}={\bf\Lambda}{\bf\Lambda}^*$, then ${\bf A}$ is normal:
  

  $${\bf A}{\bf A}^* = ({\bf U}{\bf\Lambda}{\bf U}^*)({\bf U}{\bf\Lambda}{\bf U}^*)^* =... ...*)({\bf U}{\bf\Lambda}^*{\bf U}^*) ={\bf U}{\bf\Lambda}{\bf\Lambda}^*{\bf U}^*$$

  $$= {\bf U}{\bf\Lambda}^*{\bf\Lambda}{\bf U}^* =({\bf U}{\bf\Lambda}^*{\bf U}^*)({\bf U}{\bf\Lambda}{\bf U}^*) ={\bf A}^*{\bf A}$$

  
  

• If ${\bf A}$ is normal, then it is diagonalizable by a unitary matrix. First we show any matrix ${\bf A}$ can be written as
  
  $${\bf A}={\bf B}+i{\bf C}$$
  
  
  where
  
  $${\bf B}=\frac{1}{2}({\bf A}+{\bf A}^*)={\bf B}^*,\;\;\;\;\;... ...\bf C}=-\frac{1}{2}({\bf A}-{\bf A}^*)={\bf C}^*,\;\;\;\;\;\;$$
  
  
  are both Hermitian, and diagonalizable by a unitary matrix. As ${\bf A}$ is normal, we have
  
  $$0={\bf A}{\bf A}^*-{\bf A}^*{\bf A} =({\bf B}+i{\bf C})({\... ...\bf C})({\bf B}+i{\bf C}) =2i({\bf C}{\bf B}-{\bf B}{\bf C})$$
  
  
  We see that ${\bf C}{\bf B}={\bf B}{\bf C}$, i.e., ${\bf B}$ and ${\bf C}$ commute, and they can be simultaneously diagonalized by some unitary matrix ${\bf U}$:
  
  $${\bf U}^*{\bf B}{\bf U}={\bf\Lambda}_B,\;\;\;\;\;\;\; {\bf U}^*{\bf C}{\bf U}={\bf\Lambda}_C,\;\;\;\;\;\;\;$$
  
  
  and so can ${\bf A}={\bf B}+i{\bf C}$:
  
  $${\bf U}^*{\bf A}{\bf U}={\bf U}^*({\bf B}+i{\bf C}){\bf U} ={\bf\Lambda}_B+i{\bf\Lambda}_C={\bf\Lambda}_A$$