corollary of Schur decomposition

theorem:$A\mathrm{\in }{\mathrm{C}}^{n\mathrm{\times }n}$ is a normal matrix if and only if there exists a unitary matrix $Q\in {ℂ}^{n\times n}$ such that $Q^{H}AQ=\mathrm{diag}({\lambda }_1,\mathrm{\dots },{\lambda }_n)$(the diagonal matrix) where ${}^H$ is the conjugate transpose. [GVL]

proof: Firstly we show that if there exists a unitary matrix $Q\in {ℂ}^{n\times n}$ such that $Q^{H}AQ=\mathrm{diag}({\lambda }_1,\mathrm{\dots },{\lambda }_n)$ then $A\in {ℂ}^{n\times n}$ is a normal matrix. Let $D=\mathrm{diag}({\lambda }_1,\mathrm{\dots },{\lambda }_n)$ then $A$ may be written as $A=QD{Q}^H$. Verifying that A is normal follows by the following observation $A{A}^H=QD{Q}^{H}Q{D}^H{Q}^H=QD{D}^H{Q}^H$ and $A^{H}A=Q{D}^H{Q}^{H}QD{Q}^H=Q{D}^{H}D{Q}^H$. Therefore $A$ is normal matrix because $D{D}^H=\mathrm{diag}({\lambda }_1\overline{{\lambda }_1},\mathrm{\dots },{\lambda }_n\overline{{\lambda }_n})=D^{H}D$.
Secondly we show that if $A\in {ℂ}^{n\times n}$ is a normal matrix then there exists a unitary matrix $Q\in {ℂ}^{n\times n}$ such that $Q^{H}AQ=\mathrm{diag}({\lambda }_1,\mathrm{\dots },{\lambda }_n)$. By Schur decompostion we know that there exists a $Q\in {ℂ}^{n\times n}$ such that $Q^{H}AQ=T$($T$ is an upper triangular matrix). Since $A$ is a normal matrix then $T$ is also a normal matrix. The result that $T$ is a diagonal matrix comes from showing that a normal upper triangular matrix is diagonal (see theorem for normal triangular matrices).
QED