Let \(A\in M_n(\CC)\) be normal, so \(AA^*=A^*A\). Show that \(A\) is unitarily diagonalizable.
Solution
1. By Schur triangularization, there is a unitary matrix \(U\) such that\[ T:=U^*AU \]is upper triangular.
Schur’s theorem applies to every complex square matrix. Thus \(A=UTU^*\) for some unitary \(U\) and upper-triangular \(T\).
2. The triangular matrix \(T\) is normal.
Since \(A\) is normal, \[ AA^*=A^*A. \] Conjugating by \(U\) gives \[ TT^*=T^*T. \] Thus \(T\) is normal.
3. Every upper-triangular normal complex matrix is diagonal.
We argue by induction on \(n\). The result is immediate for \(n=1\).
Write \(T=(t_{ij})\). Comparing the \((1,1)\) entries of \(TT^*\) and \(T^*T\) gives \[ |t_{11}|^2+|t_{12}|^2+\cdots+|t_{1n}|^2=|t_{11}|^2. \] Hence \[ t_{12}=\cdots=t_{1n}=0. \] Since \(T\) is upper triangular, its first column also has no entries below \(t_{11}\), so \[ T=[t_{11}]\oplus T_1 \] for an upper-triangular \((n-1)\times(n-1)\) matrix \(T_1\). The equality \(TT^*=T^*T\) then implies \[ T_1T_1^*=T_1^*T_1, \] so \(T_1\) is normal. By induction \(T_1\) is diagonal. Therefore \(T\) is diagonal.
4. Therefore \(A\) is unitarily diagonalizable.
By 1 and 3, \(T=D\) is diagonal. Hence \[ A=UDU^*, \] which is a unitary diagonalization of \(A\).